Đáp án:

`a,S={9}`

`b,S={1/2 ;3}`

`c,S={1}`

Giải thích các bước giải:

 `a,3x-11=x+7`

`<=>3x-x=7+11`

`<=>2x=18`

`<=>x=9`

Vậy phương trình có tập nghiệm `S={9}`

`b, 2x(x-3)=x-3`

`<=>2x(x-3)-(x-3)=0`

`<=>(2x-1)(x-3)=0`

`<=>`\(\left[ \begin{array}{l}2x-1=0\\x-3=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=3\end{array} \right.\) 

Vậy phương trình có tập nghiệm `S={1/2 ;3}`

`c, ĐKXĐ:` $x\ne 0,x\ne 2$

$\dfrac{x+2}{x-2}-\dfrac{5}{x}=\dfrac{8}{x^2-2x}\\
\Leftrightarrow \dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{5\left(x-2\right)}{x\left(x-2\right)}=\dfrac{8}{x\left(x-2\right)}\\ \Leftrightarrow \dfrac{x^2+2x}{x\left(x-2\right)}-\dfrac{5x-10}{x\left(x-2\right)}-\dfrac{8}{x\left(x-2\right)}=0\\ \Leftrightarrow \dfrac{\left(x^2+2x\right)-\left(5x-10\right)-8}{x\left(x-2\right)}=0\\
\Leftrightarrow \dfrac{x^2+2x-5x+10-8}{x\left(x-2\right)}=0\\
\Leftrightarrow \dfrac{x^2-3x+2}{x\left(x-2\right)}=0\\
\Leftrightarrow \dfrac{x^2-x-2x+2}{x\left(x-2\right)}=0\\
\Leftrightarrow \dfrac{x\left(x-1\right)-2\left(x-1\right)}{x\left(x-2\right)}=0\\
\Leftrightarrow \dfrac{\left(x-1\right)\left(x-2\right)}{x\left(x-2\right)}=0\\
\Leftrightarrow \dfrac{x-1}{x}=0\\
\Rightarrow x-1=0\\
\Leftrightarrow x=1\left(tm\right)$

Vậy phương trình có tập nghiệm `S={1}`

Đáp án:

 $a) 3x-11=x+7$

$⇔ 3x-x= 7+11$

$⇔ 2x=18$

$⇔ x=9$

$\text{Vậy S={9}}$

$b) 2x.(x-3)=x-3$

$⇔ 2x.(x-3)-1.(x-3)=0$

$⇔ (2x-1).(x-3)=0$

$⇔$\(\left[ \begin{array}{l}2x-1=0\\x-3=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=3\end{array} \right.\) 

$\text{Vậy S=}$   `{\frac{1}{2};3}`

$c) \dfrac{x+2}{x-2}-\dfrac{5}{x}=\dfrac{8}{x^2-2x}$ $\text{( đk: x $\neq$ 0;2)}$

$⇔ \dfrac{x.(x+2)}{x.(x-2)}-\dfrac{5.(x-2)}{x.(x-2)}=\dfrac{8}{x.(x-2)}$
$⇒x^2+2x-5x+10=8$

$⇔ x^2-3x+2=0$

$⇔ x^2-x-2x+2=0$

$⇔ x.(x-1)-2.(x-1)=0$

$⇔ (x-2).(x-1)=0$ 

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

$\text{Vậy S={1}}$