b) \(\left(2x+1\right)^2+\left(y+5\right)^4=0\)
Ta có:
\(\left\{{}\begin{matrix}\left(2x+1\right)^2\ge0\\\left(y+5\right)^4\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left(2x+1\right)^2+\left(y+5\right)^4\ge0\) \(\forall x,y.\)
\(\Rightarrow\left(2x+1\right)^2+\left(y+5\right)^4=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\\left(y+5\right)^4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x+1=0\\y+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=-1\\y=-5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=-5\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{-\frac{1}{2};-5\right\}.\)
c) \(\left|x-2\right|+\left|y+11\right|=0\)
Ta có:
\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|y+11\right|\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left|x-2\right|+\left|y+11\right|\ge0\) \(\forall x,y.\)
\(\Rightarrow\left|x-2\right|+\left|y+11\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|y+11\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2=0\\y+11=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-11\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{2;-11\right\}.\)
Chúc bạn học tốt!
