Exercice 1 :
1) Identités Remarquables :
(a+b)² = a²+2ab+b²
(a-b)² = a²-2ab+b²
(a+b)(a-b) = a²-b²
2) a) -2+9x*(-3)-21 = -2-27x-21 = -27x-23
b) (x-4)² = x²-2*x*4+4² = x²-8x+16
c) -2*(2x+1)² = -2*[(2x)²+2*2x*1+1²)] = -2*(4x²+4x+1) = -8x²-8x-2
d) (4-x)(x+6) = 4x+24-x²-6x = -x²-2x+24
3) a) 25x²-81 = (5x)²-(9)² - Cette expression est de la forme a²-b² avec a = 5x
et b = 9 -
(5x)²-(9)² = (5x+9)(5x-9)
b) 25x²-15x = 5x*5x-5x*3 - Tu remarques l'existence d'un facteur commun : 5x -
= 5x*(5x-3)
c) (x+3)(3x+5)-(x+3)(2x-7) Tu remarques l'existence d'un facteur commun : x+3
= (x+3)*[3x+5-(2x-7)]
= (x+3)*(3x+5-2x+7)
= (x+3)*(x+12)
Exercice 2 :
[tex] \frac{5}{3} - \frac{1}{6} = \frac{2*5}{2*3} - \frac{1}{6} = \frac{10}{6} - \frac{1}{6} = \frac{9}{6} = \frac{3*3}{3*2} = \frac{3}{2} [/tex]
[tex] \frac{7}{5} * \frac{-25}{14} = \frac{7*(-25)}{5*14} = - \frac{175}{70} = - \frac{35*5}{35*2} = -\frac{5}{2} [/tex]
[tex] \frac{ \frac{4}{3}-1 }{ \frac{7}{6}-2 } = \frac{ \frac{4}{3}- \frac{3}{3} }{ \frac{7}{6}- \frac{12}{6} } = \frac{ \frac{1}{3} }{- \frac{5}{6} } = \frac{1}{3} *- \frac{6}{5} =- \frac{6}{15} =- \frac{2}{5} [/tex]
[tex] -\frac{12}{7} + \frac{2}{7} / \frac{3}{5} = -\frac{12}{7} + \frac{2}{7} * \frac{5}{3} = - \frac{12}{7} + \frac{10}{21} = -\frac{36}{21} + \frac{10}{21} = - \frac{26}{21} [/tex]
