Đáp án:b)\(\left[ \begin{array}{l}x=2\\x=0\end{array} \right.\)
c)$x=0$ hoặc $x=\frac{1}{2}$ hoặc $x=-3$
d)$x=\frac{-1}{2}$
e)\(\left[ \begin{array}{l}x=3\\x=\frac{1}{3}\end{array} \right.\)
Giải thích các bước giải:
b)$5(x+3)(x-2)-3(x+5)(x-2)=0$
⇔$(x-2)[5(x+3)-3(x+5)]=0$
⇔$(x-2)[5x+15-3x-15]=0$
⇔(x-2)2x=0⇔ \(\left[ \begin{array}{l}x-2=0\\2x=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=2\\x=0\end{array} \right.\)
c)$2x^{3}+5x^{2}-3x=0$
⇔$x(2x^{2}+5x-3)=0$
⇔$x(x-\frac{1}{2})(x+3)=0$
⇔$x=0$ hoặc $x-\frac{1}{2}=0$ hoặc $x+3=0$
⇔ $x=0$ hoặc $x=\frac{1}{2}$ hoặc $x=-3$
d)$(x-1)^{2}+2(x-1)(x+2)+(x+2)^{2}=0$
⇔$(x-1+x+2)^{2}=0$
⇔$(2x+1)^{2}=0⇔ 2x+1=0⇔ x=\frac{-1}{2}$
e)$x^{2}+2x+1=4(x^{2}-2x+1)$
⇔$(x+1)^{2}=4(x-1)^{2}$
⇔$(x+1)^{2}-4(x-1)^{2}=0$
⇔$[x+1-2(x-1)][x+1+2(x-1)]=0$
⇔$(-x+3)(3x-1)=0$
⇔\(\left[ \begin{array}{l}-x+3=0\\3x-1=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=3\\x=\frac{1}{3}\end{array} \right.\)
