Question

Tìm \(x,y\) biết \(y^2-4y+14=\)\(\dfrac{20}{x^2+2x+3}\)

Answer 1

\(\Leftrightarrow\left(y-2\right)^2+10=\dfrac{20}{\left(x+1\right)^2+2}\)

Ta có: \(\left(y-2\right)^2+10\ge10\)

\(\left(x+1\right)^2+2\ge2\Rightarrow\dfrac{20}{\left(x+1\right)^2+2}\le\dfrac{20}{2}=10\)

\(\Rightarrow\left(y-2\right)^2+10\ge\dfrac{20}{\left(x+1\right)^2+2}\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}y-2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Answer 2