Question

Tìm x,y biết

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

\(2\times x+2^{x+3}=136\)

\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)

\(\left(2\times x-5\right)^{2000}+\left(3\times y+4\right)^{2002}\le0\)

Answer 1

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

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

\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)

Dấu "=" xảy ra khi:

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

đề sai câu b các câu sau áp dụng tương tự

Answer 2

c/ Vì: \(\left(x-12+y\right)^{200}+\left(x-4-x\right)^{200}=0\)

mà \(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\forall x,y\\\left(x-4-y\right)^{200}\ge0\forall x,y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-12+y=0\\x-4-y=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=12\\x-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)