ĐKXĐ: \(x\notin\left\{-\dfrac{5}{3};1\right\}\)
\(Q=\dfrac{x^2+2x+1}{3x^2+2x-5}\)
\(=\dfrac{x^2+2\cdot x\cdot1+1^2}{3x^2+5x-3x-5}\)
\(=\dfrac{\left(x+1\right)^2}{\left(3x+5\right)\left(x-1\right)}\)
Để Q>0 thì \(\dfrac{\left(x+1\right)^2}{\left(3x+5\right)\left(x-1\right)}>0\)
mà \(\left(x+1\right)^2>=0\forall x\) thỏa mãn ĐKXĐ
nên (3x+5)(x-1)>0
TH1: \(\left\{{}\begin{matrix}3x+5>0\\x-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-\dfrac{5}{3}\\x>1\end{matrix}\right.\)
=>x>1
TH2: \(\left\{{}\begin{matrix}3x+5< 0\\x-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -\dfrac{5}{3}\\x< 1\end{matrix}\right.\)
=>\(x< -\dfrac{5}{3}\)
