1, 3x-15=2x(x-5) 2, x^2-x=0 3, x^2-2x=0 4, x^2-3x=0 5, (x+1)(x+2)=(2-x)(x+2) can gap ah camon!!!

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............