Giải thích các bước giải:
1) 3x-15=2x(x-5)
⇔ 3x-15=2x²-10x
⇔ -2x²+3x+10x-15=0
⇔ -2x²+13x-15=0
⇔ 2x²-13x+15=0
⇔ 2x²-10x-3x+15=0
⇔ (2x²-10x)-(3x-15)=0
⇔ 2x(x-5)-3(x-5)=0
⇔ (2x-3)(x-5)=0
⇔ \(\left[ \begin{array}{l}2x-3=0\\x-5=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=$\frac{3}{2}$ \\x=5\end{array} \right.\)
Vậy................
2) x²-x=0
⇔ x(x-1)=0
⇔ \(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Vậy.....
3) x²-2x=0
⇔ x(x-2)=0
⇔ \(\left[ \begin{array}{l}x=0\\x-2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy....
4) x²-3x=0
⇔ x(x-3)=0
⇔ \(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy......
5) (x+1)(x+2)=(2-x)(x+2)
⇔ (x+1)(x+2)-(2-x)(x+2)=0
⇔ (x+2)[(x+1)-(2-x)]=0
⇔ (x+2)(x+1-2+x)=0
⇔ (x+2)(2x-1)=0
⇔ \(\left[ \begin{array}{l}x+2=0\\2x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-2\\x= $\frac{1}{2}$ \end{array} \right.\)
Vậy............
