Question

Cho 3 số thực x, y, z thỏa mãn \(2x^2+y^2+z^2+2xy-2xz-10x-10y+25=0\). Tìm giá trị lớn nhất của \(A=\frac{x+y+1}{z^2-z+1}\)

Answer 1

\(2x^2+y^2+z^2+2xy-2xz-10\left(x+y\right)+25=0\)

\(\Leftrightarrow\left(x+y\right)^2-10xy+25+\left(x-z\right)^2=0\)

\(\Leftrightarrow\left(x+y-5\right)^2+\left(x-z\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y-5=0\\x-z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=5\\z=x\end{matrix}\right.\)

\(A=\frac{x+y+1}{z^2-z+1}=\frac{6}{\left(z-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{6}{\frac{3}{4}}=8\)

\(A_{max}=8\) khi \(\left\{{}\begin{matrix}x+y=5\\x=z\\z=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\z=\frac{1}{2}\\y=\frac{9}{2}\end{matrix}\right.\)