ĐK: `x \ne \pm 3`
`(2(9+2x))/(x^2-9)=2/(x-3)-1/(x+3)`
`<=>2(2x+9)=2(x+3)-(x-3)`
`<=>4x+18=2x+6-x+3`
`<=>4x+18=x+9`
`<=>3x=-9`
`x=-3 (L)`
Vậy `S=∅`.
\(\dfrac{2\left(9+2x\right)}{x^2-9}=\dfrac{2}{x-3}-\dfrac{1}{x+3}\)
\(\Leftrightarrow\dfrac{18+4x}{\left(x-3\right)\left(x+3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Rightarrow18+4x-2x-6+x-3=0\)
\(\Leftrightarrow3x+9=0\)
\(\Leftrightarrow3\left(x+3\right)=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
\(S=\left\{-3\right\}\)
ĐKXĐ: x ≠ 3 ; x ≠ -3
\(\dfrac{2\left(9+2x\right)}{x^2-9}=\dfrac{2\left(x+3\right)}{x^2-9}-\dfrac{x-3}{x^2-9}\)
⇒ \(2\left(9+2x\right)=2\left(x+3\right)-1\left(x-3\right)\)
Ta có: \(2\left(9+2x\right)=2\left(x+3\right)-1\left(x-3\right)\)
\(18+4x=2x+6-x+3\)
\(4x-2x+x=6+3-18\)
\(3x=0\)
\(\Rightarrow x\) vô nghiệm
Vậy phương trình vô nghiệm
