Question

giair các phương trình sau

a)\(\dfrac{3}{1-x}-\dfrac{2}{x+2}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\)

 

Answer 1

\(\dfrac{3}{1-x}-\dfrac{2}{x+2}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\left(x\ne1;x\ne-2\right)\)

\(< =>\dfrac{-3}{x-1}-\dfrac{2}{x+2}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\)

\(< =>\dfrac{-3\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}-\dfrac{2\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=\dfrac{x+8}{\left(x-1\right)\left(x+2\right)}\)

suy ra

`-3(x+2)-2(x-1)=x+8`

`<=>-3x-6-2x+2=x+8`

`<=>-3x-2x-x=8+6-2`

`<=>-6x=12`

`<=>x=-2(ktmđk)`

Vậy phương trình vô nghiệm

Answer 2

=>-3(x+2)-2x+2=x+8

=>-3x-6-2x+2=x+8

=>-5x-4=x+8

=>-6x=12

=>x=-2(loại)