Bài 2:
a: \(\left\{{}\begin{matrix}-2x+5y=1\\3x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5y-1\\3x-2y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5}{2}y-\dfrac{1}{2}\\3\left(\dfrac{5}{2}y-\dfrac{1}{2}\right)-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y-1}{2}\\\dfrac{15}{2}y-\dfrac{3}{2}-2y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5y-1}{2}\\\dfrac{11}{2}y=4+\dfrac{3}{2}=\dfrac{11}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=\dfrac{5\cdot1-1}{2}=\dfrac{5-1}{2}=2\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}x\sqrt{3}-y=-\sqrt{3}\\3x-\sqrt{2}\cdot y=2-2\sqrt{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x\sqrt{3}+\sqrt{3}\\3x-\sqrt{2}\left(x\sqrt{3}+\sqrt{3}\right)=2-2\sqrt{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x\sqrt{3}+\sqrt{3}\\x\left(3-\sqrt{6}\right)=2-2\sqrt{6}+\sqrt{6}=2-\sqrt{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2-\sqrt{6}}{3-\sqrt{6}}=\dfrac{-\sqrt{6}}{3}\\y=\dfrac{-\sqrt{6}}{3}\cdot\sqrt{3}+\sqrt{3}=\sqrt{3}-\sqrt{2}\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}\dfrac{1}{2}x+\dfrac{3}{2}y=1\\\dfrac{3}{4}x-y=-\dfrac{1}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\y=\dfrac{3}{4}x+\dfrac{1}{8}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{3}{4}x+\dfrac{1}{8}\\x+2\left(\dfrac{3}{4}x+\dfrac{1}{8}\right)=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{4}x+\dfrac{1}{8}\\x+\dfrac{3}{2}x+\dfrac{1}{4}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{2}x=2-\dfrac{1}{4}=\dfrac{7}{4}\\y=\dfrac{3}{4}x+\dfrac{1}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{4}:\dfrac{5}{2}=\dfrac{7}{4}\cdot\dfrac{2}{5}=\dfrac{14}{20}=\dfrac{7}{10}\\y=\dfrac{3}{4}\cdot\dfrac{7}{10}+\dfrac{1}{8}=\dfrac{21}{40}+\dfrac{5}{40}=\dfrac{26}{40}=\dfrac{13}{20}\end{matrix}\right.\)
d: \(\left\{{}\begin{matrix}x\left(\sqrt{2}+1\right)-y=-2\\2x-\left(\sqrt{2}+1\right)y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x\left(\sqrt{2}+1\right)+2\\2x-\left(\sqrt{2}+1\right)\cdot\left[x\left(\sqrt{2}+1\right)+2\right]=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x\left(\sqrt{2}+1\right)+2\\2x-x\left(3+2\sqrt{2}\right)-2\left(\sqrt{2}+1\right)=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x\left(\sqrt{2}+1\right)+2\\x\left(-1-2\sqrt{2}\right)=-1+2\sqrt{2}+2=2\sqrt{2}+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-1\\y=-\left(\sqrt{2}+1\right)+2=2-\sqrt{2}-1=1-\sqrt{2}\end{matrix}\right.\)
