\(\left(x-2\right)^4+\left(2y-1\right)^{2022}< =0\)

mà \(\left(x-2\right)^4+\left(2y-1\right)^{2022}>=0\forall x,y\)

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

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

\(M=11xy^2+4xy^2=15xy^2=15\cdot2\cdot\left(\dfrac{1}{2}\right)^2=\dfrac{15}{2}\)

la

\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\2y+5=0\\4z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4}\end{matrix}\right.\)

Ta thấy : \(\left|x-2021\right|\ge0\forall x,\left|y-2022\right|\ge0\forall y\\ =>\left|x-2021\right|+\left|y-2022\right|\ge0\)

Mà theo đề : \(\left|x-2021\right|+\left|y-2022\right|\le0\)

=> \(\left\{{}\begin{matrix}x-2021=0\\y-2022=0\end{matrix}\right.=>\left(x;y\right)=\left(2021;2022\right)\)

Vì \(\left|x+\frac{8}{5}\right|\ge0;\left|2,2-2y\right|\ge0\)

=> \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\ge0\)

Mà theo đề bài \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0\)

=> \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|=0\)

=>\(\hept{\begin{cases}\left|x+\frac{8}{5}\right|=0\\\left|2,2-2y\right|=0\end{cases}}\)=>  \(\hept{\begin{cases}x+\frac{8}{5}=0\\2,2-2y=0\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{-8}{5}\\2y=2,2\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{-8}{5}\\y=1,1=\frac{11}{10}\end{cases}}\)

kb vs mk nha

Ta có: \(\left|3x-5\right|+\left(2y+5\right)^2+\left(4z-3\right)^{20}\ge0\)với \(\forall x;y;z\)

Mà \(\left|3x-5\right|+\left(2y+5\right)^2+\left(4z-3\right)^{20}\le0\)

\(\Rightarrow\left|3x-5\right|+\left(2y+5\right)^2+\left(4z-3\right)^{20}=0\)

\(\Rightarrow\hept{\begin{cases}3x-5=0\\2y+5=0\\4z-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{-5}{2}\\x=\frac{3}{4}\end{cases}}}\)

Vậy \(x=\frac{5}{3};y=\frac{-2}{5};z=\frac{3}{4}\)