a. \(\left(2x-3y\right)^2+x^2+10x+25=0\)
\(\Rightarrow\left(2x-3y\right)^2+\left(x+5\right)^2=0\) (*)
Ta có: \(\left\{{}\begin{matrix}\left(2x-3y\right)^2\ge0\\\left(x+5\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(2x-3y\right)^2+\left(x-5\right)^2\ge0\)
Nhận xét: Để tổng trên bằng 0 thì 2 số hạng phải bằng 0. Do đó:
(*) \(\Rightarrow\left\{{}\begin{matrix}\left(2x-3y\right)^2=0\\\left(x+5\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\y=-\dfrac{10}{3}\end{matrix}\right.\)
b. \(\left(3x-2y\right)^{10}+4y^2-28y+49\le0\)
\(\Rightarrow\left(3x-2y\right)^{10}+\left(2y-7\right)^2\le0\)
Ta có: \(\left\{{}\begin{matrix}\left(3x-2y\right)^{10}\ge0\\\left(2y-7\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(3x-2y\right)^{10}+\left(2y-7\right)^2\ge0\)
Mà theo đề bài: \(\left(3x-2y\right)^{10}+\left(2y-7\right)^2\le0\)
\(\Rightarrow\left(3x-2y\right)^{10}+\left(2y-7\right)^2=0\)
Lập luận tương tự như câu a, ta có:
\(\left\{{}\begin{matrix}\left(3x-2y\right)^{10}=0\\\left(2y-7\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{3}\\y=\dfrac{7}{2}\end{matrix}\right.\)
c) \(\left(5x-y\right)^2+9x^2-24x+16\le0\)
\(\Rightarrow\left(5x-y\right)^2+\left(3x-4\right)^2\le0\) (*)
Ta có \(\left\{{}\begin{matrix}\left(5x-y\right)^2\ge0\\\left(3x-4\right)^2\ge0\end{matrix}\right.\)
(*) \(\Rightarrow\left\{{}\begin{matrix}\left(5x-y\right)^2=0\\\left(3x-4\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x-y=0\\3x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{20}{3}\end{matrix}\right.\)
d) \(-y^2-6x-9-\left(3x-5y\right)^2\ge0\)
\(\Rightarrow-\left(y+3\right)^2-\left(3x-5y\right)^2\ge0\) (**)
Ta có: \(\left\{{}\begin{matrix}\left(y+3\right)^2\ge0\\\left(3x-5y\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}-\left(y+3\right)^2\le0\\-\left(3x-5y\right)^2\le0\end{matrix}\right.\)
(**) \(\Rightarrow\left\{{}\begin{matrix}\left(y+3\right)^2=0\\\left(3x-5y\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+3=0\\3x-5y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=-5\end{matrix}\right.\)
