`13/(2x^2 + x - 21 ) + 1/(2x + 7 ) =6/( x^2 - 9 )`

Đáp án:

 `x = -4`

Giải thích các bước giải:

 `13/(2x^2 + x  - 21) + 1/(2x + 7) = 6/(x^2 - 9) (x  \ne -7/2 ; x \ne \pm 3)`

`<=> 13/(2x^2 - 6x + 7x - 21) + 1/(2x + 7) = 6/((x-3)(x+3))`

`<=> 13/( 2x(x - 3) + 7(x - 3) ) + 1/(2x + 7) = 6/((x-3)(x+3))`

`<=> 13/((2x+7)(x-3)) + 1/(2x + 7) = 6/((x-3)(x+3))`

`<=> ( 13 + x - 3)/((2x + 7)(x - 3)) = 6/((x-3)(x+3))`

`<=> (x + 10)/((2x + 7)(x - 3)) = 6/((x-3)(x+3))`

`<=> ( (x + 10)(x + 3) )/((2x + 7)(x - 3)(x + 3) ) = ( 6(2x + 7) )/((2x + 7)(x - 3)(x + 3))`

`=> x^2 + 10x + 3x + 30 = 12x + 42`

`<=> x^2 + 13x + 30 - 12x - 42 = 0`

`<=> x^2 + x - 12 = 0`

`<=> x^2 + 4x - 3x - 12 = 0`

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

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

`<=>` \(\left[ \begin{array}{l}x-3=0\\x+4=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=3(L)\\x=-4(TM)\end{array} \right.\) 

Vậy `S = {-4}`