Question

\(\left(x+\dfrac{2}{3}\right)^2+\left(\dfrac{3}{2}-y\right)^8=0\)

Answer 1

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

Xét \(\left\{{}\begin{matrix}\left(x+\dfrac{2}{3}\right)^2\ge0\\\left(\dfrac{3}{2}-y\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\)\(\left\{{}\begin{matrix}\left(x+\dfrac{2}{3}\right)^2=0\\\left(\dfrac{3}{2}-y\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{3}{2}\end{matrix}\right.\)

Answer 2

\(\left(x+\dfrac{2}{3}\right)^2+\left(\dfrac{3}{2}-y\right)^8=0\)

Với mọi x thì \(\left(x+\dfrac{2}{3}\right)^2+\left(\dfrac{3}{2}-y\right)^8\ge0\)

Để \(\left(x+\dfrac{2}{3}\right)^2+\left(\dfrac{3}{2}-y\right)^2=0\) thì

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

Vậy...........