Giải hệ phương trình sau:
a, \(\left\{{}\begin{matrix}\left(\sqrt{2}+1\right)x+y=\sqrt{2}-1\\2x-\left(\sqrt{2}-1\right)y=2\sqrt{2}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\sqrt{3}x-y=1\\5x+\sqrt[]{2}y=\sqrt{3}\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}2x+y=5\\3x-2y=3\end{matrix}\right.\)
a)
\(\left\{{}\begin{matrix}\left(\sqrt{2}+1\right)x+y=\sqrt{2}-1\\2x-\left(\sqrt{2}-1\right)y=2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{2}-1\right)-\left(\sqrt{2}+1\right)x\\2x-\left(\sqrt{2}-1\right)y=2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{2}-1\right)-\left(\sqrt{2}+1\right)x\\2x-\left(\sqrt{2}-1\right)\left(\left(\sqrt{2}-1\right)-\left(\sqrt{2}+1\right)x\right)=2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{2}-1\right)-\left(\sqrt{2}+1\right)x\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(\sqrt{2}-1\right)-\left(\sqrt{2}+1\right).1\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy hệ phương trình có tập nghiệm {1;-2}
b)
\(\left\{{}\begin{matrix}\sqrt{3}x-y=1\\5x+\sqrt{2}y=\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}x-1\\5x+\sqrt{2}y=\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}x-1\\5x+\sqrt{2}\left(\sqrt{3}x-1\right)=\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}x-1\\x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}.\left(\frac{3\sqrt{3}+2\sqrt{2}}{19}\right)-1\\x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-10+2\sqrt{6}}{19}\\x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3\sqrt{3}+2\sqrt{2}}{19}\\y=\frac{-10+2\sqrt{6}}{19}\end{matrix}\right.\)
Vậy hệ phương trình có tập nghiệm \(\left\{\frac{3\sqrt{3}+2\sqrt{2}}{19};\frac{-10+2\sqrt{6}}{19}\right\}\)
c)
\(\left\{{}\begin{matrix}2x+y=5\\3x-2y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=10\\3x-2y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=13\\4x+2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{13}{7}\\4.\frac{13}{7}+2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{13}{7}\\y=\frac{9}{7}\end{matrix}\right.\)
Vậy hệ phương trình có tập nghiệm \(\left\{\frac{13}{7};\frac{9}{7}\right\}\)
