ĐKXĐ: \(x\le6;y\ge3\)
\(x^2+2y=xy+4\Leftrightarrow\left(x-2\right)\left(x-y+2\right)=0\)
✽Với \(x=2\):
Thay vào phương trình thứ 2, rút gọn ta được:
\(\left(y-3\right)\sqrt{y-3}=1\Leftrightarrow y-3=1\Leftrightarrow y=4\) (t/m)
✽Với \(y=x+2\) ta có:
\(x^2-x+3-x\sqrt{6-x}=\left(x-1\right)\sqrt{x-1}\) \(\left(1\le x\le6\right)\)
\(\Leftrightarrow2x^2-2x+6-2x\sqrt{6-x}-2\left(x-1\right)\sqrt{x-1}=0\)
\(\Leftrightarrow\left(x^2-2x\sqrt{6-x}+6-x\right)+\left(x^2-2x+1-2\left(x-1\right)\sqrt{x-1}+x-1\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{6-x}\right)^2+\left(x-1-\sqrt{x-1}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{6-x}\\x-1=\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow x=2\), suy ra \(y=4\)
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