Question

Tìm x;y biết : 

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

Answer 1

Ta có \(\left|y-1\right|+\left|y-2\right|+\left|y-3\right|+1=\left|y-1\right|+\left|y-2\right|+\left|3-y\right|+1\ge2+\left|y-2\right|+1=3+\left|y-2\right|\ge3\)

\(\dfrac{6}{\left(x-1\right)^2+2}\le\dfrac{6}{0+2}=3\)

\(\Leftrightarrow VT\le3\le VP\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\left(y-1\right)\left(3-y\right)\ge0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy PT có nghiệm \(\left(x;y\right)=\left(1;2\right)\)