Đáp án:18)\(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
19)\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
20)$x=1$ hoặc $x=-1$ hoặc $x=-3$
21)x=-3
22)$\left[ \begin{array}{l}x=1\\x=\frac{-11}{2}\end{array} \right.$
Giải thích các bước giải:
18)$(x^{2}-2x+1)-4=0$
⇔$(x-1)^{2}-2^{2}=0$
⇔$(x-1-2)(x-1+2)=0$
⇔$(x-3)(x+1)=0$⇔\(\left[ \begin{array}{l}x-3=0\\x+1=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
19)$x^{2}-x=-2x+2$
⇔$x(x-1)=-2(x-1)$
⇔$(x-1)(x+2)=0$⇔\(\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
20)$(x^{2}-1)(x+3)=0$
⇔$(x-1)(x+1)(x+3)=0$
⇔$x-1=0$ hoặc $x+1=0$ hoặc $x+3=0$
⇔ $x=1$ hoặc $x=-1$ hoặc $x=-3$
21)$(2x^{2}+1)(4x-3)=(x-12)(2x^{2}+1)$
⇔$(2x^{2}+1)(4x-3)-(x-12)(2x^{2}+1)=0$
⇔$(2x^{2}+1)(4x-3-x+12)=0$
⇔$(2x^{2}+1)(3x+9)=0$
vì $2x^{2}+1>0$ (luôn đúng)
⇒ $3x+9=0⇔ x=-3$
22)$(x-1)(5x+3)=(3x-8)(x-1)$
⇔$(x-1)(5x+3)-(3x-8)(x-1)=0$
⇔$(x-1)(5x+3-3x+8)=0$
⇔$(x-1)(2x+11)=0$⇔\(\left[ \begin{array}{l}x-1=0\\2x+11=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=\frac{-11}{2}\end{array} \right.\)
