`(3x)/(x+1)-3=x/(x+1)` (đkxđ: `x≠-1)`
`<=>(3x)/(x+1)`\(-\dfrac{3\left(x+1\right)}{x+1}=\dfrac{x}{x+1}\)
\(⇔ 3 x − 3 ( x + 1 ) = x\)
\(⇔ 3 x − 3 x − 3 = x\)
\(⇔ x = − 3 (nhận)\)
=>3x-3(x+1)=x
=>x=3x-3x-3
=>x=-3(nhận)
\(\dfrac{3x}{x+1}-3=\dfrac{x}{x+1}\)
\(ĐK:x\ne-1\)
\(\Leftrightarrow\dfrac{3x}{x+1}-\dfrac{x}{x+1}=3\)
\(\Leftrightarrow\dfrac{2x}{x+1}=3\)
\(\Leftrightarrow\dfrac{2x}{x+1}-3=0\)
\(\Leftrightarrow\dfrac{2x-3x-3}{x+1}=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
4 và 5
