Question

[ha xuan duong]

\(\dfrac{5}{x+2}-\dfrac{x-1}{x-2}=\dfrac{12}{x^2-4}+1\)
giúp vs ạ mik đng cần gấp

Answer 1

[Ngô Hải Nam]

\(\dfrac{5}{x+2}-\dfrac{x-1}{x-2}=\dfrac{12}{x^2-4}+1\left(x\ne-2;x\ne2\right)\)

\(< =>\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

suy ra

`5x-10-(x^2 +2x-x-2)=12+x^2 -4`

`<=>5x-10-x^2 -2x+x+2-12-x^2 +4=0`

`<=>-x^2 -x^2 +5x-2x+x-10+2+4=0`

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

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

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

`<=>x-2=0`

`<=>x=2(ktmđk)`

vậy phương trình vô nghiệm

Answer 2

[Nguyễn Việt Lâm]

ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow5\left(x-2\right)-\left(x-1\right)\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow5x-10-\left(x^2+x-2\right)=12+x^2-4\)

\(\Leftrightarrow-x^2+4x-8=x^2+8\)

\(\Leftrightarrow2x^2-4x+16=0\)

\(\Leftrightarrow2\left(x-1\right)^2+14=0\)

Do \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\\14>0\end{matrix}\right.\) ;\(\forall x\)

\(\Rightarrow2\left(x-1\right)^2+14>0\)

Vậy phương trình đã cho vô nghiệm