ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\frac{4}{x}\sqrt{x+1}=y\sqrt{y^2+4}\)
Nếu \(y\le0\Rightarrow\left\{{}\begin{matrix}VT>0\\VP\le0\end{matrix}\right.\) pt vô nghiệm \(\Rightarrow y>0\)
Bình phương 2 vế:
\(\frac{16}{x^2}+\frac{16}{x}=y^4+4y^2\)
\(\Leftrightarrow\frac{16}{x^2}+\frac{16}{x}+4=y^4+4y^2+4\)
\(\Leftrightarrow\left(\frac{4}{x}+2\right)^2=\left(y^2+2\right)^2\)
\(\Leftrightarrow\frac{4}{x}+2=y^2+2\Rightarrow y^2=\frac{4}{x}\)
Thay vào pt dưới:
\(\sqrt{x^2-x.\frac{4}{x}+1}+3\sqrt{x-1}=x.\frac{4}{x}\)
\(\Leftrightarrow\sqrt{x^2-3}+3\sqrt{x-1}=4\)
\(\Leftrightarrow\sqrt{x^2-3}-1+3\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2-3}+1}+\frac{3\left(x-2\right)}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow...\)
