DKXĐ: \(x\ge y;x\ge3y\)
PT (1) \(\Leftrightarrow\sqrt{x-y}+1=\sqrt{x-3y}\)
\(\Leftrightarrow x-y+1+2\sqrt{x-y}=x-3y\)
\(\Leftrightarrow2\sqrt{x-y}=-\left(2y+1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y+1\le0\\4\left(x-y\right)=\left(2y+1\right)^2\end{matrix}\right.\Rightarrow x=\frac{\left(4y^2+8y+1\right)}{4}\text{và }y\le-\frac{1}{2}\)
Thay vào PT (2) \(\Leftrightarrow\frac{7}{4}\left(4y^2+8y+1\right)-5y=19+2\sqrt{\frac{\left(4y^2+8y+1\right)}{4}-y}\)
Or: \(7\,{y}^{2}+9\,y-{\frac{69}{4}}=-(2y+1)\)
Do đó y = -5/2 hoặc 13/14 (loại)
Từ đó x = 3/2 và y =-5/2
P/s: Em không chắc lắm..
\(\left\{{}\begin{matrix}\sqrt{x-y}-\sqrt{x-3y}=-1\\7x-5y=19+2\sqrt{x-y}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-y}=a\left(a\ge0\right)\\\sqrt{x-3y}=b\left(b\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=-1\\\frac{7\left(3a^2-b^2\right)}{2}-\frac{5\left(a^2-b^2\right)}{2}=19+2a\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=a+1\\7\left(2a^2-2a-1\right)+5\left(2a+1\right)=4a+38\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=a+1\\14a^2-8a-40=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\a=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-y}=2\\\sqrt{x-3y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\y=-\frac{5}{2}\end{matrix}\right.\)
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