Đáp án:

`1)`

`(10x+3)/12=1+(6+8x)/9`

`⇔[3(10x+3)]/36=36/36+[4(6+8x)]/36`

`⇔30x+9=36+24+32x`

`⇔30x+9=32x+60`

`⇔-2x=51`

`⇔x=-51/2`

Vậy `S={-51/2}`

`2)`

`2x^(3)-5x^(2)+3x=0`

`⇔x(2x^(2)-5x+3)=0`

`⇔x(2x^(2)-2x-3x+3)=0`

`⇔x(x-1)(2x-3)=0`

$⇔\left[\begin{matrix}x=0\\ x-1=0\\ 2x-3=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=0\\ x=1\\ 2x=3\end{matrix}\right.$

$⇔\left[\begin{matrix}x=0\\ x=1\\ x=\dfrac{3}{2}\end{matrix}\right.$

Vậy `S={0;1;3/2}`

`3)`

ĐKXĐ : `x \ne 3; x \ne -1`.

`x/(2x-6)+x/(2x+2)-(2x)/[(x+1)(x-3)]=0`

`⇔x/[2(x-3)]+x/[2(x+1)]-(2x)/[(x+1)(x-3)]=0`

`⇔[x(x+1)]/[2(x-3)(x+1)]+[x(x-3)]/[2(x+1)(x-3)]-(2x.2)/[2(x+1)(x-3)]=0`

`⇔x^(2)+x+x^(2)-3x-4x=0`

`⇔2x^(2)-6x=0`

`⇔2x(x-3)=0`

$⇔\left[\begin{matrix}2x=0\\ x-3=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=0\\ x=3 (l)\end{matrix}\right.$

Vậy `S={0}`

1/ $\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\\⇔3(10x+3)=36+4(6+8x)\\⇔30x+9=36+24+32x\\⇔30x+9=60+32x\\⇔-2x=51\\⇔x=-\dfrac{51}{2}$

Vậy phương trình có tập nghiệm $S=\left\{-\dfrac{51}{2}\right\}$

2/ $2x^3-5x^2+3x=0\\⇔x(2x^2-5x+3)=0\\⇔x(2x^2-2x-3x+3)=0\\⇔x[2x(x-1)-3(x-1)]=0\\⇔x(x-1)(2x-3)=0\\⇔\left[\begin{array}{1}x=0\\x-1=0\\2x-3=0\end{array}\right.\\⇔\left[\begin{array}{1}x=0\\x=1\\2x=3\end{array}\right.\\⇔\left[\begin{array}{1}x=0\\x=2\\x=\dfrac{3}{2}\end{array}\right.$

Vậy phương trình có tập nghiệm $S=\left\{0;1;\dfrac{3}{2}\right\}$

3/ ĐK: $x\ne -1;x\ne 3$

$\dfrac{x}{2x-6}+\dfrac{x}{2x+2}-\dfrac{2x}{(x+1)(x-3)}=0\\⇔\dfrac{x}{2(x-3)}+\dfrac{x}{2(x+1)}-\dfrac{2x}{(x+1)(x-3)}=0\\⇔\dfrac{x(x+1)}{2(x+1)(x-3)}+\dfrac{x(x-3)}{2(x+1)(x-3)}-\dfrac{4x}{2(x+1)(x-3)}=0\\⇒x^2+x+x^2-3x-4x=0\\⇔2x^2-6x=0\\⇔2x(x-3)=0\\⇔\left[\begin{array}{1}x=0\\x-3=0\end{array}\right.\\⇔\left[\begin{array}{1}x=0(TM)\\x=3(KTM)\end{array}\right.$

Vậy phương trình có tập nghiệm $S=\{0\}$