Toán Lớp 8: Tìm x : a) 3(x -1) (x +1) = x+1 b) 9x(x -5) = (5 -x)(5+x)

Question

Toán Lớp 8:  Tìm x : a) 3(x -1) (x +1) = x+1 b) 9x(x -5) = (5 -x)(5+x)

kymmailien · Answer

a) 3(x -1)(x +1) = x+1     &hArr; 3(x -1)(x +1) - (x+1) = 0     &hArr; (3x - 3)(x+1) - (x+1) = 0     &hArr; ( x+1 )( 3x - 3 - 1 ) = 0     &hArr; ( x+1 )( 3x - 4 ) = 0     &hArr;  ( left[  begin{array}{l}x+1=0  3x-4=0 end{array}  right. )      &hArr;  ( left[  begin{array}{l}x=-1  x=4/3 end{array}  right. )      Vậy S = { -1; $ frac{4}{3}$ }      b) 9x(x -5) = (5 -x)(5+x)    &hArr;  9x(x -5) - (5 -x)(5+x) = 0     &hArr; 9x(x -5) + (x-5)(5+x) = 0     &hArr; (x-5)(9x + 5 + x) = 0     &hArr; (x-5)(10x +5 )= 0     &hArr;  ( left[  begin{array}{l}x-5=0  10x+5=0 end{array}  right. )      &hArr;  ( left[  begin{array}{l}x=5  x=-1/2 end{array}  right. )      Vậy S = { 5; $ frac{-1}{2}$ }

mytamhoang · Answer

Giải đáp+Lời giải và giải thích chi tiết:     a,   $3(x-1)(x+1)=x+1$ $&hArr;(3x-3)(x+1)-(x+1)=0$ $&hArr;(3x-3-1)(x+1)=0$ $&hArr;(3x-4)(x+1)=0$ $&hArr; left[ begin{matrix}x+1=0  3x-4=0 end{matrix} right.$ $&hArr; left[ begin{matrix}x=-1  3x=4 end{matrix} right.$   $&hArr; left[ begin{matrix}x=-1  x= dfrac{4}{3} end{matrix} right.$   Vậy S={-1; frac{4}{3}}   b,   $9x(x-5)=(5-x)(5+x)$ $&hArr;9x(x-5)-(5-x)(5+x)=0$ $&hArr;9x(x-5)+(x-5)(5+x)=0$ $&hArr;(x-5)(9x+5+x)=0$ $&hArr;(x-5)(10x+5)=0$ $&hArr;5(x-5)(2x+1)=0$ $&hArr; left[ begin{matrix}x-5=0  2x+1=0 end{matrix} right.$   $&hArr; left[ begin{matrix}x=5  2x=-1 end{matrix} right.$   $&hArr; left[ begin{matrix}x=5  x= dfrac{-1}{2} end{matrix} right.$   Vậy S={5; frac{-1}{2}}