Toán Lớp 8: Giúp mk vs ạ :9x^2=1-(3x+1)(2x-9)

Question

Toán Lớp 8:  Giúp mk vs ạ :9x^2=1-(3x+1)(2x-9)

truonganhthi · Answer

Giải đáp+Lời giải và giải thích chi tiết:    9x&sup2;=1-(3x+1)(2x-9)    &hArr;9x&sup2;-1=-(3x+1)(2x-9)   &hArr; 9x&sup2;-1=-(6x&sup2;-27x+2x-9)   &hArr; 9x&sup2;-1=-(6x&sup2;-25x-9)   &hArr; 9x&sup2;-1=-6x&sup2;+25x+9   &hArr; 9x&sup2;-1+6x&sup2;-25x-9=0   &hArr;15x&sup2;-10-25x=0   &hArr;5(3x&sup2;-2-5x)=0   &hArr;3x&sup2;-2-5x=0   &hArr;3x&sup2;-5x-2=0   &hArr;3x&sup2;-6x+x-2=0( T&aacute;ch hạng tử -5x th&agrave;nh -6x+x)   &hArr;(3x&sup2;-6x)+(x-2)=0 ( Nh&oacute;m hạng tử)   &hArr;3x(x-2)+(x-2)=0( Đặt nh&acirc;n tử chung)   &hArr;(x-2)(3x+1)=0   &hArr; ( left[  begin{array}{l}x-2=0  3x+1=0 end{array}  right. )  &hArr;  ( left[  begin{array}{l}x=2  3x=-1 end{array}  right. )  &hArr; ( left[  begin{array}{l}x=2  x=-1/3  end{array}  right. )    Vậy tập nghiệm của phương tr&igrave;nh l&agrave; S={ 2; $ frac{-1}{3}$ }

tonhu12 · Answer

9x^2 = 1 - (3x+1)(2x-9) <=>9x^2 -1 = - (6x^2 -27x + 2x - 9) <=>9x^2 -1 = -6x^2+ 25x +9 <=>9x^2 -1 + 6x^2 - 25x - 9=0 <=>15x^2 - 25x - 10 =0 <=>5(3x^2 - 5x -2)=0 <=>3x^2- 5x-2=0 <=>3x^2 - 6x + x -2=0 <=>3x(x-2) + (x-2)=0 <=>(3x+1)(x-2)=0 <=> ( left[ begin{array}{l}3x+1=0 x-2=0 end{array} right. ) <=> ( left[ begin{array}{l}x= frac{-1}{3} x=2 end{array} right. ) Vậy x in { -1/3 ;2}