Grade 10 / Remarkable identities 96 exercises (100% corrected)

a
abcda×ca×db×cb×c abFig.1abFig.2abFig.3 ChingQuizz : 11 exercises available for Quizz assessment : 1. Reminders E.11438 Definition: " double distributivity "is the action of ex-panding two factors involving sums/and subtractions. Proposition: Let a , b , c , d be four numbers. We have the identity: Expand, then reduce the following expressions : a (3 x + 1)(3 + 4 x ) b (3 x + 2)(5 x + 4) c (2 x +5)(3+2 x ) d (2 x + 1)(3 x + 2) E.11439 Expand, then reduce the following expressions : E.11440 Expand, then reduce the following expressions : E.11441 Expand, then reduce the following expressions : E.9686 Using the method of your choice, solve the following equations : a 3 x 2 + x = 0 b (3 x + 1) 2 = 3 x + 1 E.11608 Solve the following equations : a 2 x − 1 3 =5 x +1 b x − 4 3 = x − 2 c 2 x +1 6 − 1 − x 2 = x E.474 Solve the following equations : a ( x + 1)(2 − x ) = (2 x − 4)(5 x − 3) b (2 x + 1)(3 x + 4) − (3 x + 1)(2 x + 4) = 0 E.11613 Expand the following expressions : a 2(3 x − 1)(2 − x ) b 5( x + 3)(5 − 2 x ) 2. Introduction by double distributivity E.11442 1 To complete the expansions and simplifications, copy and fill in the blanks : a x + 3 2 = x + 3 ( x + 3 = : : : b x + 10 2 = x + 10 ( x + 10 = : : : 2 Copy and complete the sentences : The numerical term of the expanded expression is . . . . . . of the numerical term of the squared factor. The coefficient of the term in " x "of the expanded ex-pression is . . . . . . of the numerical term of the squared factor. E.11443 1 To complete the developments, copy and fill in the blanks : a x − 4 2 = x − 4 x − 4 = : : : b x − 5 2 = x − 5 x − 5 = : : : 2 With b ∈ R and considering the developed expression of x − b 2 : The numerical term has the value : . . . The term in " x "a has the expression :. . . E.11444 To complete the developments, copy and fill in the blanks : a x + 2 x − 2 = : : : b x + 10 x − 10 = : : : 3. Introduction by graphic identification E.8175 In this exercise, consider a square of side a + b where a and b are two positive real numbers ( a;b ∈ 0 ; + ∞ ) . 1 For each of the figures below, give the area of the hatched domain : https://chingmath.fr chapExoCorrec/11438 sacados/11438 abcda×ca×db×cb×c chapExoCorrec/11439 sacados/11439 chapExoCorrec/11440 sacados/11440 chapExoCorrec/11441 sacados/11441 chapExoCorrec/9686 sacados/9686 chapExoCorrec/11608 sacados/11608 chapExoCorrec/474 sacados/474 chapExoCorrec/11613 sacados/11613 chapExoCorrec/11442 sacados/11442 chapExoCorrec/11443 sacados/11443 chapExoCorrec/11444 sacados/11444 chapExoCorrec/8175 sacados/8175 abFig.1abFig.2abFig.3
abFig.1abFig.2abFig.3abFig.4 abFig.1abFig.2aabbFig.3aabbFig.4 abaabb 2 From the expressions below, give the two answers for ex-pressing the area of the square : a a + b 2 b a 2 + b 2 c a 2 + 2 ab + b 2 d a 2 − 2 ab + b 2 E.8185 Let a and b be two strictly positive real numbers. Consider the four representations of the same square of side a below : 1 a Express the area of each of the hatched parts using the numbers a and b . b Which part of this figure admits as area the expression : a − b 2 +2 ab − b 2 2 Justify identity: a − b 2 = a 2 − 2 ab + b 2 E.8186 Let a and b be two strictly positive real numbers such that b<a . Consider below a square of side a (Figs. 1 and 2) and a rectangle (Figs. 3 and 4) : 1 Express in terms of a and b the areas of the hatched domains above. 2 a What can be said about the areas of the shaded do-mains shown below? b Justify the identity: a 2 − b 2 = a + b a − b 4. Development and remarkable identity E.8179 1 Establish each of the identities below : a 3 x + 5 2 = 3 x 2 + 2 × 3 x × 5 + 5 2 b 4 x + 3 2 = 4 x 2 + 2 × 4 x × 3 + 3 2 2 Establish each of the identities below : a 2 x − 1 2 = 2 x 2 − 2 × 2 x × 1 + 1 2 b 3 − 6 x 2 = 3 2 − 2 × 3 × 6 x + 6 x 2 3 Establish each of the identities below : a x + 2 x − 2 = x 2 − 2 2 b 4 x + 5 4 x − 5 = 4 x 2 − 5 2 5. Developing a remarkable identity E.8180 Complete the table below : a + b 2 a b a 2 b 2 2 ab a 2 + 2 ab + b 2 3 x +2 2 4 x +1 2 5 x +1 2 E.8181 Complete the table below : a − b 2 a b a 2 2 ab b 2 a 2 − 2 ab + b 2 x − 5 2 2 x − 4 2 4 x − 3 2 E.8182 Complete the table below : a + b a − b a b a 2 b 2 a 2 − b 2 2 x +5 2 x − 5 x +4 x − 4 4 x +3 4 x − 3 E.8176 Expand the following expressions : a ( x + 1) 2 b (2 x + 3) 2 c ( x + 6) 2 d (5 x + 1) 2 e (3 x + 3) 2 f ( a + b ) 2 E.438 Expand the following expressions : a (2 x + 3) 2 b (3 x − 2)(3 x + 2) c (5 x − 6) 2 https://chingmath.fr chapExoCorrec/8185 sacados/8185 abFig.1abFig.2abFig.3abFig.4 chapExoCorrec/8186 sacados/8186 abFig.1abFig.2aabbFig.3aabbFig.4 abaabb chapExoCorrec/8179 sacados/8179 chapExoCorrec/8180 sacados/8180 chapExoCorrec/8181 sacados/8181 chapExoCorrec/8182 sacados/8182 chapExoCorrec/8176 sacados/8176 chapExoCorrec/438 sacados/438
E.8177 Expand the following expressions : a ( x − 2) 2 b ( x − 3) 2 c (3 x − 1) 2 d (5 x − 1) 2 e (3 x − 2) 2 f ( a − b ) 2 E.11528 Expand and reduce the following expressions : E.681 Copy on your copy and complete so that the equalities are true : a (3 x + : : : ) 2 = : : : + 18 x + : : : b (3 x − : : : )(3 x + : : : ) = 9 x 2 − 9 4 c ( x + : : : )( : : : − 1) = 3 x 2 + : : : − 2 d ( : : : − : : : ) 2 = : : : − 24 x + 9 E.8174 Complete the blanks below to obtain . a 2 x + 4 2 = 4 x 2 + 16 x + : : : b 3 x + 1 2 = : : : + 6 x + 1 c x − 2 2 = : : : − 4 x + 4 d 4 + 5 x 2 = 16 + 40 x + : : : e x − 3 2 = x 2 − 6 x + : : : E.9687 Expand and reduce the following expressions : a 1 2 x + 3 4 2 b 4 3 x − 6 5 2 c 1 2 x + 1 2 1 2 x − 1 2 6. Develop E.691 Give the expanded and reduced forms of the following literal expressions : a ( x + 1) 2 + (2 x − 1) 2 b 2 x + 1 + (4 x − 3) 2 c 3 + (5 + x ) 2 d ( x + 1)( x − 1) (2 x − 3) E.11529 Expand and reduce the following expressions : a 2 x − 2 2 + 3 2 x − 1 b 6 x + 3 2 − 4 3 x + 2 7. Factorize a remarkable identity E.678 Consider the following literal expres-sions : a 81 x 2 + 80 x + 25 b 4 x 2 − 12 x + 9 c 16 x 2 − 32 x − 16 d 36 − 4 x 2 1 Remarkable identities can be used to write the following factorizations : a 2 + 2 · ab + b 2 = ( a + b ) 2 a 2 − 2 · ab + b 2 = ( a − b ) 2 a 2 − b 2 = ( a + b )( a − b ) Identifying, if possible, each of the proposed expressions with one of the remarkable identities, complete the table below : a b 2 · ab a b c d 2 Which of the proposed expressions can be factorized? We will then give their factorized form. E.9690 Consider the following literal expres-sions : a 25 x 2 + 20 x + 4 b 9 x 2 + 18 x + 9 c 4 x 2 − 12 x + 9 d 25 x 2 − 16 1 Remarkable identities can be used to perform the follow-ing factorizations : a 2 + 2 · ab + b 2 = ( a + b ) 2 a 2 − 2 · ab + b 2 = ( a − b ) 2 a 2 − b 2 = ( a + b )( a − b ) Identifying, if possible, each of the proposed expressions with one of the remarkable identities, complete the table below : a b 2 · ab a b c d 2 Which of the following expressions are factorized? We will then give their factorized form. https://chingmath.fr chapExoCorrec/8177 sacados/8177 chapExoCorrec/11528 sacados/11528 chapExoCorrec/681 sacados/681 chapExoCorrec/8174 sacados/8174 chapExoCorrec/9687 sacados/9687 chapExoCorrec/691 sacados/691 chapExoCorrec/11529 sacados/11529 chapExoCorrec/678 sacados/678 chapExoCorrec/9690 sacados/9690
E.5175 1 Of the three expressions below only one has been ob-tained by developing a remarkable identity? Which one is it? Specify the starting expression : a 4 x 2 + 6 x + 9 b 4 x 2 + 24 x + 9 c 4 x 2 + 12 x + 9 2 Same question with the expressions : a x 2 − 64 x + 64 b x 2 − 16 x + 64 c x 2 − 8 x + 64 3 Same question with the expressions : a 9 x 2 +15 x +25 b 9 x 2 +30 x +25 c 9 x 2 +6 x +25 E.9676 Factor each of the following expres-sions : a 9 x 2 − 12 x + 4 b x 2 + 2 x + 1 E.2237 Factor each of the following literal expressions : a 25 x 2 − 40 x + 16 b 81 x 2 − 90 x + 25 c 49 x 2 + 84 x + 36 d 100 x 2 − 25 E.2236 Factor each of the following literal expressions : a x 2 − 16 b x 2 − 10 x + 25 c x 2 − 2 x + 1 d x 2 + 14 x + 49 E.9664 Factor the following algebraic ex-pressions : a x 2 − 20 x + 100 b x 2 − 4 x + 4 c x 2 − 9 d x 2 + 12 x + 36 E.9663 Factor the following literal expres-sions : a x 2 + 2 3 · x + 1 9 b 1 4 · x 2 − 1 9 c 1 9 x 2 − 2 15 x + 1 25 d 1 4 x 2 − 1 3 x + 1 9 E.702 Factor, if possible , the following literal expressions, highlighting your approach : a 4 x 2 − 24 x + 9 b 9 + 24 x − 16 x 2 c 64 x 2 − 9 d 9 x 2 + 30 x + 25 e 9 x 2 + 12 x + 4 f 16 x 2 + 20 x + 25 E.2238 Factor, if possible , the following literal expressions : a 25 x 2 − 50 x + 25 b 4 x 2 + 1 c 100 x 2 + 140 x + 49 d 4 x 2 + 24 x + 9 E.9669 Factor the following expressions : b 4 x 4 − 9 a x 4 − 4 x 2 + 4 b 9 x 4 − 12 x 2 + 4 E.5903 Factor the following expressions : a − x 2 − 4 x − 4 b − x 2 + 6 x − 9 c − 9 x 2 + 12 x − 4 d − 25 x 2 + 20 x − 4 8. Factoring: a little further E.700 Factorize the following expressions. No particular justification is required : b ( x + 2) 2 − 9 b 25 x 2 − 9 − (5 x + 3)(5 − x ) E.674 Factor the following expressions ; no explanation is required : a (3 x + 1) 2 − (2 − 2 x ) 2 b 25 x 2 − 4 − (5 x + 2)(5 x − 4) E.689 Without justification, factor the fol-lowing expressions : a 25 x 2 − 36 + (2 − x )(5 x − 6) b (2 x + 5) 2 − (1 − x ) 2 E.449 Factor the following expressions : a (2 x − 8)(7 x + 1) − 16 + x 2 b 18 x 2 − 24 x + 8 + (3 x − 2)(2 − x ) E.8514 Establish the following factoriza-tion : 3 2 x + 1 + x − 1 2 = x + 2 2 E.8515 Determine the value of the reals a and b performing the factorization : 2 x + 1 4 x + 3 + x + 1 2 = ax + b 2 9. Equation: development of remarkable identities E.11463 Solve the equations : a x − 2 2 − x 2 + 4 = 0 b 2 x + 3 2 − 4 x 2 − 1 = 0 E.11462 Solve the equations : a x + 4 4 x + 5 − 2 x + 5 2 = 0 b x − 3 2 − x − 5 x + 3 = 0 https://chingmath.fr chapExoCorrec/5175 sacados/5175 chapExoCorrec/9676 sacados/9676 chapExoCorrec/2237 sacados/2237 chapExoCorrec/2236 sacados/2236 chapExoCorrec/9664 sacados/9664 chapExoCorrec/9663 sacados/9663 chapExoCorrec/702 sacados/702 chapExoCorrec/2238 sacados/2238 chapExoCorrec/9669 sacados/9669 chapExoCorrec/5903 sacados/5903 chapExoCorrec/700 sacados/700 chapExoCorrec/674 sacados/674 chapExoCorrec/689 sacados/689 chapExoCorrec/449 sacados/449 chapExoCorrec/8514 sacados/8514 chapExoCorrec/8515 sacados/8515 chapExoCorrec/11463 sacados/11463 chapExoCorrec/11462 sacados/11462
10. Equation: reminders about the zero product equation E.11461 Solve the equations : a x 2 − 4 x = 0 b 5 x 2 + 3 x = 0 Hint: We will factor the left side to obtain a zero product equation. 11. Equation: factorization of remarkable identities E.5329 Solve the equations : a 4 x 2 + 12 x + 9 = 0 b x 2 − 10 x + 25 = 0 c 4 x 2 − 9 = 0 Hint: We will factor using the remarkable identities to ob-tain a zero product equation E.9689 Solve the equations : a 9 x 2 − 12 x + 4 = 0 b 25 x 2 − 9 = 0 c 4 x 2 + 20 x + 25 = 0 Hint: We will factor using the remarkable identities to ob-tain a zero product equation E.11459 Solve the equations : a 81 x 2 − 18 x = − 1 b x 2 + 5 x − 5 = 3 x − 6 Hint: We will factor using the remarkable identities to ob-tain a zero product equation E.11530 Solve the equations : a 16 x 2 + 9 = 24 x b 10 x + 1 = − 25 x 2 Hint: We will factor using the remarkable identities to ob-tain a zero product equation. E.9688 Solve the equations below. To do this, use factorization to obtain a null product equation. a 10 x 2 + 30 x + 30 = x 2 + 5 b x 2 + 1 = 2 x c 16 x 2 + 4 x + 3 = 4 x + 7 E.9675 Solve the equations : a x + 3 2 − 4 x 2 b ( x + 1) 2 − (2 x − 3) 2 = 0 E.9680 Consider the algebraic expression : B =(5 x − 7) 2 − 3 2 1 Determine the factorized form of the expression B . 2 Find a value of x for which B =0 . 12. Equation: expansion, remarkable identity E.11464 Solve the equations : a x + 2 4 x + 5 = x + 1 b 3 x + 1 2 + (2 x + 3)(8 x + 5) = 0 Hint: Expand the expressions, then factorize using a re-markable identity to obtain a zero product equation. E.11465 Solve the equations : a 3 x + 1 2 + (2 x − 2)(8 x + 5) = 0 b (4 x − 1) 2 + ( x + 4)(9 x + 2) = 0 Hint: Expand the expressions, then factorize using a re-markable identity to obtain a zero product equation. E.11466 Solve the equations : a (3 x + 7) 2 − ( x + 4)(5 x + 10) = 0 b (8 x − 2) 2 − (3 x − 1)(5 x + 5) = 0 Hint: Expand the expressions, then factorize using a re-markable identity to obtain a zero product equation. 13. Equation: common factor, remarkable identity E.832 Modify the given equations to obtain zero product equations, then solve them : b 16 x 2 + 24 x + 9 = (3 x − 2) 2 E.11460 Solve the following equations : a ( x + 1)(2 x − 3) = 4 x 2 − 9 b ( x + 1)(2 x − 3) = x 2 − 1 https://chingmath.fr chapExoCorrec/11461 sacados/11461 chapExoCorrec/5329 sacados/5329 chapExoCorrec/9689 sacados/9689 chapExoCorrec/11459 sacados/11459 chapExoCorrec/11530 sacados/11530 chapExoCorrec/9688 sacados/9688 chapExoCorrec/9675 sacados/9675 chapExoCorrec/9680 sacados/9680 chapExoCorrec/11464 sacados/11464 chapExoCorrec/11465 sacados/11465 chapExoCorrec/11466 sacados/11466 chapExoCorrec/832 sacados/832 chapExoCorrec/11460 sacados/11460
ABCD4x−25−x ABCDEFGHI2xx4cm4cm E.8933 Consider the expression : A = 3 x + 1 3 x + 8 + 3 x + 1 x − 2 − 9 x 2 − 1 1 Put the expression A into the form of a product of two factors. 2 For what values of x does this expression cancel? 14. Rational expression E.8507 Establish the following identity: x +2 x +1 − x +2 2 x +3 = x +2 2 x +1 2 x +3 E.8512 1 Determine the values of the reals a and b realizing the identity: 2 x + 3 x − 2 x + 3 3 x + 3 = ax + b 2 x 3 x + 3 2 Determine the values of the reals c and d realizing the identity: 5 x + 2 2 x + 1 − 3 x − 2 3 x + 1 = cx + d 2 2 x + 1 3 x + 1 E.8513 Consider the expression A defined by: A = 2 x + 3 x + 1 + x + 1 Write the expression A as a quotient whose numerator is the square of a first-degree polynomial. 15. Problems E.5681 The expanded formof remarkable identities is shown below : ( a + b ) 2 = a 2 + 2 ab + b 2 ; ( a − b ) 2 = a 2 − 2 ab + b 2 ( a + b )( a − b ) = a 2 − b 2 Use these remarkable identities to determine by mental calcu-lation the value of the calculations below : a 21 2 b 29 2 c 21 × 19 d 34 × 26 E.5263 Consider the rectan-gle ABCD shown opposite, whose di-mensions, depending on an indeterminate value x , are 5 − x and 4 x − 2 expressed in centimeters. Determine the possible values of x so that the area of ABCD , expressed in cm 2 , is equal to the perimeter of ABDC , ex-pressed in cm . E.8190 Consider the shaded figure below and note its area A : (measurements are in centimetres) It is composed of : square AEFG , of two rectangles ABCD and CIFH . Determine the value(s) of x so that the area A has the value 7 cm 2 Any trace of research or initiative will be taken into account in the assessment. E.3376 In this exercise, any trace of research, however incomplete, or initiative, however unsuccessful, will be taken into account in the assessment. Aissa asserts : ˇFor any natural integer n , the expression 2 n 2 − 6 n +4 is always the square of an integer.ı Is he right? https://chingmath.fr chapExoCorrec/8933 sacados/8933 chapExoCorrec/8507 sacados/8507 chapExoCorrec/8512 sacados/8512 chapExoCorrec/8513 sacados/8513 chapExoCorrec/5681 sacados/5681 chapExoCorrec/5263 sacados/5263 ABCD4x−25−x chapExoCorrec/8190 sacados/8190 ABCDEFGHI2xx4cm4cm chapExoCorrec/3376 sacados/3376
x110x−1ABC ABCDE2cm4cmxcm5cm 16cmxx10cm E.459 Let x be a real number strictly greater than 9 . Determine the value(s) of x for which triangle ABC is a right triangle. E.3760 In this exercise, any trace of research, however incomplete, or initative, however unsuccess-ful, will be taken into account in the assessment. Anatole states : ˇFor any natural number n , the expression n 2 − 24 n +144 is always non-zero.ı Is he right? E.2937 In the plane, consider two triangles ABC and EDC rectangles in A and D re-spectively such that the points A , C , D are aligned. We note x the distance, in centimeters, separating the points A and C . 1 Express as a function of x the length of segment [ BC ] . 2 a Solve the equation : x 2 +4=(5 − x ) 2 +16 b Deduce the length of segment [ AC ] so that lengths CB and CE are equal. Justify your approach. E.2864 A rectangular box without a lid is to be made in the pattern below. The lengths are expressed in cm . 1 a When the box is built, the number x will represent which dimension? Length, width or height? b What values can the variable x take in this problem? c Give the expression for the volume V as a function of the value of x . 2 In this question, we investigate for what values of ˇ x ı, this box has a volume equal to 144 cm 3 : a Determine the value of the reals of a and b verifying the following factorization : 4 x 3 − 52 x 2 + 160 x − 144 = ( a · x + b )(2 x − 4) 2 b Deduce the values of x for which V ( x ) has the value 144. E.10433 In this exercise, any trace of re-search, however incomplete, or initiative, however unsuccess-ful, will be taken into account in the assessment. Thomas states : ˇFor any natural number n , the expression 6 n 2 − 18 n +16 is always the square of an integer.ı Is he right? E.4646 A farmer has 200 meters of fenc-ing. Using the whole fence, he wishes to enclose the largest rectangular-shaped part of his field. We note x and y the respective length and width of this rect-angular part. 1 Establish identity: x · y = 1 4 · ( x + y ) 2 − 1 4 · ( x − y ) 2 2 a What relationship must x and y verify in order for the area of its field to be maximal? b Deduce the maximum area of its field. 16. Study of functions and remarkable identities E.444 Consider the two functions f and g defined by: f ( x ) = x 2 ; g ( x ) = 2 x − 1 1 Using your calculator, give the abscissas of the points of intersection of the two curves C f and C g representative of the functions f and g . 2 a Find the result of the previous question by solving the equation : x 2 = 2 x − 1 b Determine the coordinates of the intersection point of the curves C f and C g . https://chingmath.fr chapExoCorrec/459 sacados/459 x110x−1ABC chapExoCorrec/3760 sacados/3760 Centres Etrangers II Juin 2009 chapExoCorrec/2937 sacados/2937 ABCDE2cm4cmxcm5cm chapExoCorrec/2864 sacados/2864 16cmxx10cm chapExoCorrec/10433 sacados/10433 chapExoCorrec/4646 sacados/4646 chapExoCorrec/444 sacados/444
-4-3-2-12I2JOCf -4-3-2-1I-2-123JO -12345678IJOCgCf E.9721 Consider the two functions f and g defined by: f ( x ) = − 2 x − 2 ; g ( x ) = 4 x 2 + 6 x + 2 In the plane provided with a reference frame O ; I ; J or-thonormal, we note C f and C g respectively, the representa-tive curves of the functions f and g . The curve C g is given below : 1 Solve the equation : f ( x )= g ( x ) 2 Give, if they exist, the coordinates of the intersection points of the curves C f and C g . E.4450 Consider the function f defined on R whose image of a number x is given by the relation: f ( x ) = − 1 2 (4 x + 7)( x + 2) + x 2 + 4 x + 7 In the reference frame O ; I ; J orthogonal below are repre-sented the curve C f of the function f : 1 Graphically answer the following questions : a Determine the image of the number − 3 by the function f . Justify your answer. b Determine the set of antecedents of the number 0 by the function f . Justify your answer. 2 a Expand expression : − 1 2 (4 x + 7)( x + 2) + x 2 + 4 x + 7 b Deduce the set of solutions to the equation : f ( x )=0 . 3 a Factorize the expression x 2 +4 x +4 . b Deduce the factorization of the expression : − 2 x − 7 2 ( x + 2) + x 2 + 4 x + 4 c Deduce the set of solutions to the equation : f ( x ) = 3 E.4561 Consider the two functions f and g defined on − 1 ; + ∞ whose images of a number x are defined by the relations : f ( x ) = 5 x + 3 5 x + 5 ; g ( x ) = x 2 x 2 + 1 In the reference frame O ; I ; J , are drawn the curves C f and C g representative of the functions f and g : 1 Establish the following equality: g ( x ) − f ( x ) = 2 x 2 − 5 x − 3 ( x 2 + 1)(5 x + 5) 2 a Justify that 3 is a solution of the equation : f ( x ) = g ( x ) . b Determine the values of the reals a and b verifying the following equality: 2 x 2 − 5 x − 3 = ( x − 3)( a · x + b ) 3 Deduce the set of solutions to the equation : f ( x ) = g ( x ) https://chingmath.fr chapExoCorrec/9721 sacados/9721 -4-3-2-12I2JOCf chapExoCorrec/4450 sacados/4450 -4-3-2-1I-2-123JO chapExoCorrec/4561 sacados/4561 fichierPlus/4561/diapoCorrection.pdf -12345678IJOCgCf
-4-3-2-12345678910I-3-2-12345JOCfCgCg -3-2-12I-4-2246JOCf E.1016 Consider the functions f and g whose images of a number x are defined by: f ( x ) = 6 x + 4 x 2 + 2 ; g ( x ) = 3 x − 2 3 x − 1 In the frame O ; I ; J below, the representative curves of the functions f and g are given : The following questions will be answered by algebraic calcu-lations ; the representations are there to verify your results : 1 a Justify that the function f is defined on R . b Determine the image of − 5 2 number by the function f . c Determine the antecedents of the number 2 by the func-tion f . 2 a Give the definition set of the function g . b Determine by the function g the set of antecedents of − 1 . 3 Determine the coordinates of the intersection points of the curves C f and C g . E.4445 In the ( O ; I ; J ) orthogonal shown below, the curve C f is the graphical representation of a func-tion f defined on R \{− 1 } : 1 a Determine, graphically, the image of the number 1 by the function f . Justify your answer. b Solve, graphically, the equation f ( x )=0 . 2 The image of a number x by the function f is given by the relation: f ( x ) = 2 x 3 + 5 x 2 + x − 2 x + 1 a Justify that the function f admits as definition set R \{− 1 } . b Justify, by calculation, the value of the image of the number 1 . c Establish the following equalities: 2 x 3 + 5 x 2 + x − 2 x + 1 = 2 x 2 + 3 x − 2 = ( x + 2)(2 x − 1) d Solve, by calculation, the equation f ( x )=0 . 17. Remarkable entities and square roots E.762 Consider the two numbers below : A = 1 + 3 ; B = 3 − 3 Perform the following calculations : a A 2 b B 2 c A × B d A + B Give the results in the form a + b c where a , b and c are integers. E.735 Perform the following calculations using the remarkable identities and give the result in the sim-plest form possible : a 5 + 2 2 b 3 − 5 3 + 5 c 9 − 2 2 d 2 − 3 2 + 3 E.731 Expand the calculations below and give their results in the form a + b c , where a , b , c are integers with c as small as possible : a 3 − 2 2 b 5 − 3 2 − 5 + 3 2 E.253 Expand and simplify each of the ex-pressions below : a 2 2 + 3 3 2 b 2 + 5 + 6 2 5 − 6 c 5 − 2 7 1 − 7 d 1 + 2 3 2 https://chingmath.fr chapExoCorrec/1016 sacados/1016 -4-3-2-12345678910I-3-2-12345JOCfCgCg chapExoCorrec/4445 sacados/4445 -3-2-12I-4-2246JOCf chapExoCorrec/762 sacados/762 chapExoCorrec/735 sacados/735 chapExoCorrec/731 sacados/731 chapExoCorrec/253 sacados/253
E.740 Consider the expression : E = 7+1 2 + 7 − 1 2 1 After developing the squares, show that E is an integer. 2 Deduce the nature of a triangle whose sides measure, re-spectively, in centimeters, 7+1 , 7 − 1 and 4; justify your answer. E.293 Let a and b be two numbers such that a 0 and b 0 : 1 Expand : ( a − b ) 2 2 What is the sign of : ( a − b ) 2 3 Deduct : a + b 2 a × b E.263 1 a Prove the following equality: 1 − 2 2 2 =9 − 4 2 b Use this to derive a simplified expression for 9 − 4 2 . 2 Prove the following equality: 37+12 7 =3+2 7 18. Systems of non-linear equations and remarkable identities E.8197 1 Solve the system : x + y = 2 x × y = 1 2 Solve the system : x + 2 y = 4 x × y = 2 3 Solve the system : x + 3 y = − 6 x × y = 3 E.8198 1 Solve the system : 4 x + 49 y = 28 x × y = 1 2 Solve the system : 8 x + 9 y = 24 x × y = 2 3 Solve the system : 50 x + 6 y = − 60 x × y = 3 E.8199 Determine the dimensions of a rect-angle such that : its perimeter measures 16 m its area measure 16 m 2 19. Share E.706 Factor the following expressions. a 4 x 2 + 4 x + 1 b 9 x 2 − 24 x + 16 c 4 x 2 − 81 d ( x + 1)( x − 3) + 2( x + 1) E.695 Factor the following literal expres-sions, highlighting your approach : a 9 x 2 + 30 x + 25 b 64 x 2 − 49 c ( x + 1) 2 − (2 − 3 x )( x + 1) d (2 x − 1)(3 x + 4) + (2 x − 1) E.3763 Factor each of the following expres-sions : a 49 x 2 − 42 x + 9 b 16 x 2 − 1 c (5 x + 2)(3 − 2 x ) − (5 x + 2)( x + 1) d (9 x − 4) 2 − (9 x − 4) 20. Unclassified exercises E.8395 Indication : In this exercise, we establish Brahmagupta’s identity For all real numbers a , b , c , d , establish the identity below : a 2 + b 2 c 2 + d 2 = ac + bd 2 + ad − bc 2 E.836 Expand and reduce the following expressions : a ( x + 1) 2 b 2 − 2 x 2 + 2 x Factor the following expressions : c 9 x 2 − 12 x + 4 d 2 x 2 − 1 Solve the following equation : https://chingmath.fr chapExoCorrec/740 sacados/740 Centres etrangers - Juin 2004 - 3 points chapExoCorrec/293 sacados/293 chapExoCorrec/263 sacados/263 chapExoCorrec/8197 sacados/8197 chapExoCorrec/8198 sacados/8198 chapExoCorrec/8199 sacados/8199 chapExoCorrec/706 sacados/706 chapExoCorrec/695 sacados/695 chapExoCorrec/3763 sacados/3763 chapExoCorrec/8395 sacados/8395 chapExoCorrec/836 sacados/836
e ( x − 1)(2 x + 5) = 0 E.9684 Solve the equations : a (2 x − 3)(5 x + 4) = (2 x − 3)(3 − 2 x ) b (2 x + 1) 2 = (2 x + 1)(3 x − 1) https://chingmath.fr chapExoCorrec/9684 sacados/9684