Question

Giải phương trình:x+y+z+4=2√x-2 +4√y-3 +6√z-5

Answer 1

\(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

\(\Leftrightarrow x+y+z+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\)

\(\Leftrightarrow\left(x-2\sqrt{x-2}-1\right)+\left(y-4\sqrt{y-3}+1\right)+\left(z-6\sqrt{z-5}+4\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)

Ta thấy: \(VT\ge0\forall x,y,z\). Dấu "=" xảy ra khi và chỉ khi

\(\left\{\begin{matrix}\left(\sqrt{x-2}-1\right)^2=0\\\left(\sqrt{y-3}-2\right)^2=0\\\left(\sqrt{z-5}-3\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)