\(\Leftrightarrow2\left(x^2+1\right)-2x\sqrt{x^2+1}=5\)
\(\Leftrightarrow x^2+1-2x\sqrt{x^2+1}+x^2=4\)
\(\Leftrightarrow\left(\sqrt{x^2+1}-x\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}-x=2\\\sqrt{x^2+1}-x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x+2\left(x\ge-2\right)\\\sqrt{x^2+1}=x-2\left(x\ge2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+4x+4\\x^2+1=x^2-4x+4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{3}{4}\\x=\frac{3}{4}< 2\left(l\right)\end{matrix}\right.\)
ĐKXĐ: \(x\ge\frac{3}{2}\)
\(\Leftrightarrow\sqrt{5x-1}+\sqrt{2x-3}=\sqrt{3x-2}\)
\(\Leftrightarrow7x-4+2\sqrt{\left(5x-1\right)\left(2x-3\right)}=3x-2\)
\(\Leftrightarrow\sqrt{10x^2-17x+3}=1-2x\)
Do \(x\ge\frac{3}{2}\Rightarrow1-2x< 0\)
Phương trình vô nghiệm
