Question

( x + 3 )²⁰²²  + ( \(\sqrt{y-2}\) - 1 )²⁰²³ = 0

mình cần gấp mong mọi người giúp ạ!!

Answer 1

\(\left(x+3\right)^{2022}+\left(\sqrt{y-2}-1\right)^{2023}=0\)    \(\left(ĐKXĐ: y\ge2\right)\)

Xét \(\left(x+3\right)^{2022}\ge0\forall x\)

\(\Rightarrow\left(\sqrt{y-2}-1\right)^{2023}\le0\)

\(\Leftrightarrow\sqrt{y-2}-1\le0\)

\(\Leftrightarrow\sqrt{y-2}\le1\) 

\(\Leftrightarrow y-2\le1\)

\(\Rightarrow y\le3\)

\(\Rightarrow2\le y\le3\) mà \(y\in Z\)

\(\Rightarrow\left\{{}\begin{matrix}y=2\Leftrightarrow x=-2\\y=3\Leftrightarrow x=-3\end{matrix}\right.\)