Question

cho hệ phương trình: \(\left\{{}\begin{matrix}2x-ay=b\\ax+by=1\end{matrix}\right.\) tìm a, b để hệ phương trình có nghiệm \(x=\sqrt{2};y=\sqrt{3}\)

Answer 1

Thay \(x=\sqrt{2};y=\sqrt{3}\)ta có:

\(\left\{{}\begin{matrix}2\sqrt{2}-a\sqrt{3}=b\\a\sqrt{2}+b\sqrt{3}=1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}a\sqrt{3}=2\sqrt{2}-b\\a\sqrt{2}+b\sqrt{3}=1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}a=\dfrac{2\sqrt{2}-b}{\sqrt{3}}\\\sqrt{2}\cdot\dfrac{2\sqrt{2}-b}{\sqrt{3}}+b\sqrt{3}=1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}a=\dfrac{2\sqrt{2}-b}{\sqrt{3}}\\4-b\sqrt{2}+3b=\sqrt{3}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}b\left(\sqrt{2}-3\right)=4-\sqrt{3}\\a=\dfrac{2\sqrt{2}-b}{\sqrt{3}}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}b=\dfrac{4-\sqrt{3}}{\sqrt{2}-3}\\a=\dfrac{2\sqrt{2}-\dfrac{4-\sqrt{3}}{\sqrt{2}-3}}{\sqrt{3}}=\dfrac{4-6\sqrt{2}-4+\sqrt{3}}{\sqrt{3}\left(\sqrt{2}-3\right)}=\dfrac{\sqrt{3}-6\sqrt{2}}{\sqrt{3}\left(\sqrt{2}-3\right)}=\dfrac{1-2\sqrt{6}}{\sqrt{2}-3}\end{matrix}\right.\)