`P=(x+2)/(x+3)-5/(x^2+x-6)+1/(2-x)`
`=(x+2)/(x+3)-5/((x-2)(x+3))-1/(x-2)`
`=((x+2)(x-2)-5-(x+3))/((x-2)(x+3))`
`=(x^2-4-5-x-3)/((x-2)(x+3))`
`=(x^2-x-12)/((x-2)(x+3))`
`=((x+3)(x-4))/((x-2)(x+3))`
`=(x-4)/(x-2)`
`P=1/3`
`<=>(x-4)/(x-2)=1/3`
`<=>3(x-4)=x-2`
`<=>3x-12=x-2`
`<=>3x-x-12+2=0`
`<=>2x-10=0`
`<=>x-5=0`
`<=>x=5.`
Vậy `x=5` thì `P=1/3`
mình sửa bài nhé mình nhầm x + 2 ở mẫu dòng 2
\(P=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)ĐK : \(x\ne-3;2\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
Ta có \(P=\dfrac{1}{3}\Rightarrow\dfrac{x-4}{x-2}=\dfrac{1}{3}\Rightarrow3x-12=x-2\Leftrightarrow2x=10\Leftrightarrow x=5\)
