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

1, 3x-15= 2x( x-5)

⇔ 3x -15 = 2x² -10x

⇔ 3x -2x² +10x -15 = 0

⇔ -2x² +13x -15 = 0

⇔ -2x² +10x +3x -15 = 0

⇔ -2x(x -5) +3(x-5) = 0

⇔ (x-5).(-2x +3) = 0

TH1: x-5 = 0 ⇔ x = 5

TH2: -2x+3 = 0 ⇔ x= 3/2

Vậy S= {5; 3/2}

2, 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 S ={0; 1}

3, x(x-2)=0

Bạn làm tương tự như câu 2, nha . Mk cho kết quả thôi nhé!

Vậy S ={0; 2}

4, x(x-3)=0

Câu này cx thế nhé!

Vậy S ={0; 3}

5, x²+2x+x+2=2x+4-x²-2x

  ⇔ x²+2x+x+2-2x-4+x²+2x=0

  ⇔2x²+3x - 2  =0

⇔2x²-x+4x -2 =0

⇔ x.(2x-1)+2(2x-1)=0

⇔\(\left[ \begin{array}{l}2x-1=0\\x+2=0\end{array} \right.\)

⇔\(\left[ \begin{array}{l}x=1/2\\x=-2\end{array} \right.\) 

Vậy S = {1/2; -2}