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

Question

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

hoaiphuong85 · Answer

Lời giải và giải thích chi tiết: a, 3x(x-1)+x-1=0 <=>(x-1)(3x+1)=0 <=> ( left[ begin{array}{l}x-1=0 3x+1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=1 x= dfrac{-1}{3} end{array} right. ) Vậy S={1;-1/3} b,2(x+3)-x^2-3x=0 <=>2(x+3)-(x^2-3x)=0 <=>2(x+3)-x(x+3)=0 <=>(x+3)(2-x)=0 <=> ( left[ begin{array}{l}x+3=0 2-x=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-3 x=2 end{array} right. ) Vậy S={-3;2}

buianhtuan · Answer

$ color{pink}{ text{sakurangoc2007}}$         Xin hay nhất nha!! ????    a)  3x(x-1)+x-1=0      &hArr; 3x (x - 1) + (x - 1) = 0     &hArr; (3x + 1) (x - 1) = 0     &hArr;  ( left[  begin{array}{l}3x+1=0  x-1=0 end{array}  right. )      &hArr;  ( left[  begin{array}{l}3x=-1  x=1 end{array}  right. )      &hArr;  ( left[  begin{array}{l}x= dfrac{-1}{3}  x=1 end{array}  right. )                               Vậy x &isin; {$ dfrac{-1}{3}$ ; 1}     b)  2 (x + 3)- $x^2$ - 3x = 0   &hArr; 2 (x + 3) - x (x + 3) = 0   &hArr; (2 - x) (x + 3) = 0   &hArr;  ( left[  begin{array}{l}2 - x=0  x+3=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=2  x=-3 end{array}  right. )                            Vậy x  &isin; {2; -3}