Toán Lớp 8: tìm x a)3x(x-1)-2(1-x)=0 b)(x-3)^2=(2x+5)^2

Question

Toán Lớp 8:  tìm x a)3x(x-1)-2(1-x)=0 b)(x-3)^2=(2x+5)^2

dongoctuan · Answer

gửi bạn   a) 3x(x-1)-2(1-x)=0   &rArr;3x(x-1)+2(x-1)=0   &rArr;(3x+2) (x-1)=0   &rArr;$ left  { {{(3x+2)=0}  atop {(x-1)=0}}  right.$   &rArr;x=$ frac{-2}{3}$    x=1   vậy   x=$ frac{-2}{3}$    x=1   b)  (x-3)&sup2;=(2x+5)&sup2;     &rArr;(x-3)&sup2;-(2x+5)&sup2;=0     &rArr;((x-3)-(2x+5))((x-3)+(2x+5))=0     &rArr;(x-3-2x-5)(x-3+2x+5)=0     &rArr;(-x-8)(3x+2)=0     &rArr;(-x-8)=0     (3x+2)=0     &rArr;x=-8     x=$ frac{-2}{3}$      vậy     x=-8     x=$ frac{-2}{3}$     ch&uacute;c bạn học tốt

truonganhthi · Answer

a) 3x(x - 1) - 2(1 - x) = 0   => 3x(x - 1) + 2(x - 1) = 0   => (3x + 2)( x - 1) = 0   => ( left[  begin{array}{l}3x + 2= 0  x - 1 = 0 end{array}  right. )    => ( left[  begin{array}{l}3x=-2  x=1 end{array}  right. )    => ( left[  begin{array}{l}x= frac{-2}{3}  x=1 end{array}  right. )    Vậy x &isin; { -2/3 ; 1 }   b) (x - 3)^{2} = (2x + 5)^{2}   => (x - 3)^{2} - (2x + 5)^{2} = 0   => ( x - 3 - 2x - 5)( x - 3 + 2x + 5) = 0   => (-8 - x)( 3x + 2) = 0   => ( left[  begin{array}{l}-8 - x=0  3x + 2 = 0 end{array}  right. )   => ( left[  begin{array}{l}x=-8  3x=-2 end{array}  right. )   => ( left[  begin{array}{l}x=-8  x= frac{-2}{3} end{array}  right. )   Vậy x &isin; { -8 ; -2/3 }   Bạn tham khảo!