ĐKXĐ: \(\dfrac{1}{2}\le x;y\le1\)
Trừ pt trên cho dưới ta được:
\(\sqrt{2x-1}-\sqrt{2y-1}+\sqrt{1-y}-\sqrt{1-x}=0\)
\(\Leftrightarrow\dfrac{2x-2y}{\sqrt{2x-1}+\sqrt{2y-1}}+\dfrac{-y+x}{\sqrt{1-y}+\sqrt{1-x}}=0\)
\(\Leftrightarrow\left(x-y\right)\left(\dfrac{2}{\sqrt{2x-1}+\sqrt{2y-1}}+\dfrac{1}{\sqrt{1-y}+\sqrt{1-x}}\right)=0\)
\(\Leftrightarrow x-y=0\Rightarrow x=y\) (phần ngoặc phía sau luôn dương)
Thay vào pt đầu:
\(\sqrt{2x-1}+\sqrt{1-x}=1\Leftrightarrow x+2\sqrt{\left(2x-1\right)\left(1-x\right)}=1\)
\(\Leftrightarrow2\sqrt{-2x^2+3x-1}=1-x\Leftrightarrow4\left(-2x^2+3x-1\right)=\left(1-x\right)^2\)
\(\Leftrightarrow9x^2-14x+5=0\) \(\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=\dfrac{5}{9}\Rightarrow y=\dfrac{5}{9}\end{matrix}\right.\) (đều t/m)
Vậy hệ đã cho có 2 cặp nghiệm \(\left(x;y\right)=\left(\dfrac{5}{9};\dfrac{5}{9}\right);\left(1;1\right)\)
