Réponse:
Bonsoir
113 a
(x²+1)²=(3x²-1)²
(x²+1)²-(3x²-1)²=0
[(x²+1)+(3x²-1)][(x²+1)-(3x²-1)]=0
4x²(-2x²+2) = 0
4x² = 0 ou -2x²+2=0
x²=0 ou x²=1
x=0 ou x = -1 ou x =1
S={-1; 0; 1}
113b
1/(3-2x)² = 4/(x-1)²
1/(3-2x)² - 4/(x-1)² = 0
[(x+1)²-(3-2x)²]/[(3-2x)(x+1)] = 0
(x+1)²-(3-2x)² = 0
(x+1+3-2x)(x+1-3+2x)=0
(4-x)(3x-2)=0
4-x=0 ou 3x-2=0
x=4 ou x = ⅔
S={⅔; 4}
