Đặt \(\sqrt{x^2+1}=a\left(a\ge0\right)\)
=> \(x^2+1=a^2\)
Có a2+5x=(x+5)a
<=>\(a^2+5x-ax-5a=0\)
<=> \(a\left(a-x\right)-5\left(a-x\right)=0\)
<=> \(\left(a-5\right)\left(a-x\right)=0\)
=> \(\left[{}\begin{matrix}a=5\\a=x\end{matrix}\right.\)
Tự làm nốt nha:)
\(\Leftrightarrow4x^2-4xy+y^2+z^2+9y^2+1-6yz-2z+6y+z^2-4z+4\le0\)
\(\Leftrightarrow\left(2x-y\right)^2+\left(z-3y-1\right)^2+\left(z-2\right)^2\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\z-3y-1=0\\z-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{6}\\y=\frac{1}{3}\\z=2\end{matrix}\right.\)
