Question

\(\left\{{}\begin{matrix}(\sqrt{5}+2)x+y=3-\sqrt{5}\\\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\)

Answer 1

HPT \(\Leftrightarrow\left\{{}\begin{matrix}\left(2+\sqrt{5}\right)x+y=3-\sqrt{5}\\x=2y-6+2\sqrt{5}\end{matrix}\right.\)

\(\Rightarrow\left(2+\sqrt{5}\right)\left(2y-6+2\sqrt{5}\right)+y=3-\sqrt{5}\)

\(\Leftrightarrow4y+2\sqrt{5}y-12-6\sqrt{5}+4\sqrt{5}+10+y=3-\sqrt{5}\)

\(\Leftrightarrow4y+2\sqrt{5}y-12-6\sqrt{5}+4\sqrt{5}+10+y-3+\sqrt{5}=0\)

\(\Leftrightarrow y\left(5+2\sqrt{5}\right)-12-6\sqrt{5}+4\sqrt{5}+10-3+\sqrt{5}=0\)

\(\Leftrightarrow y\left(2\sqrt{5}+5\right)=5+\sqrt{5}\)

\(\Leftrightarrow y=3-\sqrt{5}\)

- Thay lại vào PT ( II ) ta được : \(x=2\left(3-\sqrt{5}\right)-6+2\sqrt{5}=0\)

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