Question

[Mai Vu]

\(\hept{\begin{cases}\sqrt{\left(x-1\right)\left(3-y\right)}+\sqrt{\left(y-1\right)\left(3-x\right)}=2\\x^2+y^2-4x-4y+7=0\end{cases}}\)2

Answer 1

[Tran Le Khanh Linh]

Ta có \(\sqrt{\left(y-1\right)\left(x-3\right)}\le\frac{x-1+3-y}{2}=1+\frac{x}{2}-\frac{y}{2}\)

\(\sqrt{\left(y-1\right)\left(3-x\right)}\le\frac{y-1+3-x}{2}=1-\frac{x}{2}+\frac{y}{2}\)

Nên \(2=\sqrt{\left(x-1\right)\left(3-y\right)}+\sqrt{\left(y-1\right)\left(3-x\right)}\le1+\frac{x}{2}-\frac{y}{2}+1-\frac{x}{2}+\frac{y}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=3-y\\y-1=3-x\end{cases}\Leftrightarrow x+y=4}\)

\(\Rightarrow x^2+y^2-4x-4y+7=0\Leftrightarrow\left(x+y\right)^2-2xy-4\left(x+y\right)+7=0\)

\(\Leftrightarrow xy=\frac{7}{2}\)

Answer 2

[Phí Yến Nhi]

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