Question

\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)

Answer 1

ĐK: `x \ne \pm 2`

`12/(x^2-4)-(x+1)/(x-2)+(x+7)/(x+2)=0`

`<=>12-(x+1)(x+2)+(x+7)(x-2)=0`

`<=> 2x-4=0`

`<=> x=2 (L)`

Vậy `S={∅}`.

Answer 2

\(x\ne\pm2\)

\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)

=> \(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow12-x^2-x-2x-2+x^2+7x-2x-14=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow x=2\) (ktm)
Vậy pt vô nghiệm.

 

Answer 3

ĐKXĐ: \(x\ne-2;x\ne2\).

Ta có \(PT\Leftrightarrow\dfrac{12}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow2x=4\Leftrightarrow x=2\) (loại)

Vậy PT đã cho vô nghiệm