Question

Tìm x;y biết  \(\dfrac{6}{\left(x-2\right)^2+3}=\left|y-1\right|+2\) 

Answer 1

Do \(\left(x-2\right)^2\ge0;\forall x\Rightarrow\dfrac{6}{\left(x-2\right)^2+3}\le\dfrac{6}{0+3}=2\) (1)

\(\left|y-1\right|\ge0;\forall y\Rightarrow\left|y-1\right|+2\ge2\) (2)

Từ (1); (2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{6}{\left(x-2\right)^2+3}=2\\\left|y-1\right|+2=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left|y-1\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)