Đáp án:
Giải thích các bước giải:
1) $x(x+2)-x^{2}+1=0_{}$
⇔ $x^{2}-x-^{2}+1=0_{}$
⇔ $x+1=0_{}$
⇔ $x=-1_{}$
2) $4x(x-1)-3(x+2)=0_{}$
⇔ $4x(x-2)+3x-6=0_{}$
⇔ $4x(x-2)+3(x-2)=0_{}$
⇔ $(x-2)(4x+3)=0_{}$
⇔ \(\left[ \begin{array}{l}x-2=0\\4x+3=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=2\\x=-\frac{3}{4}\end{array} \right.\)
3) $x(x+2)-3(x+2)=0_{}$
⇔ $(x-3)(x+2)=0_{}$
⇔ \(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
4) $8x(x-5)-2x+10=0_{}$
⇔ $8x(x-5)-2(x-5)=0_{}$
⇔ $(8x-2)(x-5)=0_{}$
⇔ \(\left[ \begin{array}{l}8x-2=0\\x-5=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=\frac{1}{4}\\x=5\end{array} \right.\)
5) $x(2x-3)-2(3-2x)=0_{}$
⇔ $x(2x-3)+2(2x-3)=0_{}$
⇔ $(x+2)(2x-3)=0_{}$
⇔ \(\left[ \begin{array}{l}x+2=0\\2x-3=0\end{array} \right.\) \(\left[ \begin{array}{l}x=-2\\x=\frac{3}{2}\end{array} \right.\)
