ĐKXĐ:\(-\frac{3}{2}\le x;y\le4\)
Trừ theo vế:\(\left(\sqrt{2x-3}-\sqrt{2y-3}\right)+\left(\sqrt{4-y}-\sqrt{4-x}\right)=0\)
\(\Leftrightarrow\frac{2\left(x-y\right)}{\sqrt{2x+3}+\sqrt{2y+3}}+\frac{x-y}{\sqrt{4-y}+\sqrt{4-x}}=0\)
\(\Leftrightarrow\left(x-y\right)\left(\frac{2}{\sqrt{2x+3}+\sqrt{2y+3}}+\frac{1}{\sqrt{4-x}+\sqrt{4-y}}\right)=0\)
\(\Leftrightarrow x=y\)(vì ĐKXĐ=>cả ngoặc bên phải dương)
Thay x=y vào phương trình đầu:
\(\sqrt{2x+3}+\sqrt{4-x}=4\)
\(\Leftrightarrow2x+3+4-x+2\sqrt{\left(2x+3\right)\left(4-x\right)}=16\)
\(\Leftrightarrow2\sqrt{\left(2x+3\right)\left(4-x\right)}=9-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}9-x\ge0\\4\left(2x+3\right)\left(4-x\right)=\left(9-x\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le9\\9x^2-38x+33=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x=3\\x=\frac{11}{9}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\rightarrow y=3\\x=\frac{11}{9}\rightarrow y=\frac{11}{9}\end{matrix}\right.\left(tmđkxđ\right)\)Vậy hệ có nghiệm (x;y)\(\in\left\{\left(3;3\right);\left(\frac{11}{9};\frac{11}{9}\right)\right\}\)
