Question

\(\left\{{}\begin{matrix}\sqrt{3}x-y=1\\x+y=\sqrt{3}\end{matrix}\right.\)

giải hệ PT

Answer 1

Giải:

Ta có: \(\left\{{}\begin{matrix}\sqrt{3}x-y=1\\x+y=\sqrt{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}x-1\\x+y=\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow x+\sqrt{3}x-1=\sqrt{3}\)

\(\Leftrightarrow x\left(1+\sqrt{3}\right)-\left(1+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(1+\sqrt{3}\right)=0\)

\(\Leftrightarrow x-1=0\) ( do \(1+\sqrt{3}>0\) )

\(\Leftrightarrow x=1\)

\(\Leftrightarrow y=\sqrt{3}-1\)

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