\(\Leftrightarrow log_2\left(\dfrac{x^2+x}{y\left(x+4\right)}\right)+\dfrac{x^2+x}{\sqrt{y}}-\dfrac{y\left(x+4\right)}{\sqrt{y}}=log_3\left(\dfrac{y\left(x+4\right)}{x^2+x}\right)\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x^2+x}{\sqrt{y}}=a>0\\\dfrac{y\left(x+4\right)}{\sqrt{y}}=b>0\end{matrix}\right.\)
\(\Rightarrow log_2\left(\dfrac{a}{b}\right)+a-b=log_3\left(\dfrac{b}{a}\right)\)
\(\Leftrightarrow log_2a+log_3a+a=log_2b+log_3b+b\)
Xét hàm \(f\left(t\right)=log_2t+log_3t+t\) với \(t>0\)
\(\Rightarrow f'\left(t\right)=\dfrac{ln2}{t}+\dfrac{ln3}{t}+1>0;\forall t>0\)
\(\Rightarrow f\left(t\right)\) đồng biến \(\Rightarrow a=b\)
\(\Rightarrow x^2+x=y\left(x+4\right)\Rightarrow y=\dfrac{x^2+x}{x+4}\)
\(y=\dfrac{x^2+x-12+12}{x+4}=x-3+\dfrac{12}{x+4}\)
\(\Rightarrow x+4\inƯ\left(12\right)=\left\{1;2;3;4;6;12\right\}\)
\(\Rightarrow y\in\left\{1;6\right\}\)
