Grade 10 / Remarkable identities 96 exercises corrected

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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 quizzChap/1080 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
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 E.10562 Establish : ( x + a )( x + a + 2) + 1 = ( x + a + 1) 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 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 sacados/10562 chapExoCorrec/11463 sacados/11463
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 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 https://chingmath.fr chapExoCorrec/11462 sacados/11462 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
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 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. https://chingmath.fr chapExoCorrec/832 sacados/832 chapExoCorrec/11460 sacados/11460 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
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? 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/3376 sacados/3376 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
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
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