ĐKXĐ: \(-\dfrac{3}{2}\le x;y\le4\)
\(\Leftrightarrow\sqrt{2x+3}-\sqrt{2y+3}+\sqrt{4-y}-\sqrt{4-x}=0\)
\(\Leftrightarrow\dfrac{2\left(x-y\right)}{\sqrt{2x+3}+\sqrt{2y+3}}+\dfrac{x-y}{\sqrt{4-y}+\sqrt{4-x}}=0\)
\(\Leftrightarrow\left(x-y\right)\left(\dfrac{2}{\sqrt{2x+3}+\sqrt{2y+3}}+\dfrac{1}{\sqrt{4-y}+\sqrt{4-x}}\right)=0\)
\(\Leftrightarrow x-y=0\) (do \(\dfrac{2}{\sqrt{2x+3}+\sqrt{2y+3}}+\dfrac{1}{\sqrt{4-y}+\sqrt{4-x}}>0\))
\(\Rightarrow x=y\)
Thay vào pt trên:
\(\sqrt{2x+3}+\sqrt{4-x}=4\Leftrightarrow2x+3+4-x+2\sqrt{\left(2x+3\right)\left(4-x\right)}=16\)
\(\Leftrightarrow2\sqrt{12+5x-2x^2}=9-x\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\4\left(12+5x-2x^2\right)=\left(9-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\9x^2-38x+33=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3;y=3\\x=\dfrac{11}{9};y=\dfrac{11}{9}\end{matrix}\right.\)
