Đáp án + Giải thích các bước giải:

`1.`

`a)` `(x-3)/x=2-(x-3)/(x+3) ( x ne 0 , xne -3)`

`<=> [(x-3)(x+3)]/[x(x+3)] = [2x(x+3)]/[x(x+3)]-[x(x-3)]/[x(x+3)]`

`=>(x-3)(x+3)=2x(x+3)-x(x-3)`

`<=>x^2-9=2x^2+6x-x^2+3x`

`<=>x^2-9-2x^2-6x+x^2-3x=0`

`<=>-9x-9=0`

`<=>-9x=9`

`<=>x=-1(tm)`

Vậy `S={-1}`

`b)` `|x^2-1|=|-3|`

`<=>|x^2-1|=3`

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

`<=>`\(\left[ \begin{array}{l}x^2=4\\x^2=-2(L)\end{array} \right.\)

`<=>x^2=4<=>x=pm2`

Vậy `S={pm2}`

`c)` `3x^2-2x-16=0`

`<=>3x^2+6x-8x-16=0`

`<=>3x(x+2)-8(x+2)=0`

`<=>(x+2)(3x-8)=0`

`<=>`\(\left[ \begin{array}{l}x+2=0\\3x-8=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=-2\\x=\dfrac{8}{3}\end{array} \right.\)

Vậy `S={-2;8/3}`

$a)$ĐKXĐ: $x\neq0; x\neq-3$

$\frac{x-3}{x}=2-\frac{x-3}{x+3}$

`<=>`$\frac{(x-3)(x+3)}{x(x+3)}=\frac{2x(x+3)}{x(x+3)}-\frac{x(x-3)}{x(x+3)}$

`=>`$(x-3)(x+3)=2x(x+3)-x(x-3)$

`<=>`$x^{2}-9=2x^{2}+6x-x^{2}+3x$

`<=>`$9x=-9$

`<=>`$x=-1 (t/m)$

Vậy $S=\{-1\}$

$b)|x^{2}-1|=|-3|$

`<=>`$|x^{2}-1|=3$

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

`<=>`\(\left[ \begin{array}{l}x^{2}=4\\x^{2}=-2 (loại)\end{array} \right.\) 

`<=>`$x=±2$

Vậy $S=\{±2\}$

$c)3x^{2}-2x-16=0$

`<=>`$(x+2)(3x-8)=0$

`<=>`\(\left[ \begin{array}{l}x+2=0\\3x-8=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-2\\x=\frac{8}{3}\end{array} \right.\) 

Vậy $S=\{-2; \frac{8}{3}\}$