Question

\(\left\{{}\begin{matrix}\left(18x^2+18x+18y-17\right)\left(12x^2-12xy-1\right)=0\\3x+4y=0\end{matrix}\right.\)

Answer 1

Từ phương trình \(\left(2\right)\): \(3x+4y=0\Leftrightarrow y=-\dfrac{3}{4}x\)

Thế vào phương trình \(\left(1\right)\) ta được:

\(\left(18x^2+\dfrac{9}{2}x-17\right)\left(21x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3\pm\sqrt{553}}{24}\\x=\pm\dfrac{\sqrt{21}}{21}\end{matrix}\right.\)

\(x=\dfrac{-3+\sqrt{553}}{24}\Rightarrow y=\dfrac{3-\sqrt{553}}{32}\)

\(x=\dfrac{-3-\sqrt{553}}{24}\Rightarrow y=\dfrac{3+\sqrt{553}}{32}\)

\(x=\dfrac{\sqrt{21}}{21}\Rightarrow y=-\dfrac{\sqrt{21}}{28}\)

\(x=-\dfrac{\sqrt{21}}{21}\Rightarrow y=\dfrac{\sqrt{21}}{28}\)

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