`(3 - x)/(x - 2) + (x + 1)/(x + 2) = (3x)/(x^2 - 4)` (đkxđ: `x \ne \pm 2`)

`<=> {(3 - x)(x + 2)}/{(x - 2)(x + 2)} + {(x + 1)(x - 2)}/{(x - 2)(x + 2)} = (3x)/{(x + 2)(x - 2)}`

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

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

`<=> - 3x = - 4`

`<=> x = 4/3` $\text{(nhận)}$

Vậy `S = {4/3}`

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`2x^2 - x - 3 = 0`

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

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

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

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

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

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

Vậy `S = {- 1; 3/2}`

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Đáp án`+`Giải thích các bước giải:

`a)`

`{3 - x}/{x - 2} + {x + 1}/{x + 2} = {3x}/{x^2 - 4}`    (ĐKXĐ: x ne +- 2`)

`<=> {3 - x}/{x - 2} + {x + 1}/{x + 2} = {3x}/{(x - 2)(x + 2)}`

`<=> {(3 - x)(x + 2)}/{(x - 2)(x + 2)} + {(x + 1)(x - 2)}/{(x - 2)(x + 2)} = {3x}/{(x - 2)(x + 2)}`\

`=> (3 - x)(x + 2) + (x + 1)(x - 2) = 3x`

`<=> (3x + 6 - x^2 - 2x) + (x^2 - 2x + x - 2) = 3x`

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

`<=> - 3x + 4 = 0`

`<=> - 3x = - 4`

`<=> x = 4/3` (Nhận)

Vậy `S = {4/3}`

`b)`

`2x^2 - x - 3 = 0`

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

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

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

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

`<=> [(2x - 3 = 0),(x + 1 = 0):}`

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

Vậy `S = {3/2 ; - 1}`