Question

\(\left\{{}\begin{matrix}\dfrac{1}{1+\left(x-y\right)^2}=z+4\\\sqrt{z+3}+2x=8\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(z\ge-3\Rightarrow z+4\ge1\)

Mà \(\dfrac{1}{1+\left(x-y\right)^2}\le1\)

\(\Rightarrow\dfrac{1}{1+\left(x-y\right)^2}\le z+4\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}z=-3\\x=y\end{matrix}\right.\)

Thay vào pt dưới \(\Rightarrow\left(x;y;z\right)=\left(4;4;-3\right)\)