Bài 5:
a: \(\left\{{}\begin{matrix}4x-3y+5\left(x-y\right)=1\\2x-4\left(2y-1\right)=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-3y+5x-5y=1\\2x-8y+4=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}9x-8y=1\\2x-8y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=4\\2x-8y=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{4}{7}\\8y=2x+3=2\cdot\dfrac{4}{7}+3=\dfrac{8}{7}+3=\dfrac{29}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{4}{7}\\y=\dfrac{29}{56}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}3\left(x-7\right)-6\left(x-y+1\right)=0\\4\left(x-1\right)+2\left(x-2y+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-21-6x+6y-6=0\\4x-4+2x-4y+14=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-3x+6y-27=0\\6x-4y+10=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-3x+6y=27\\6x-4y=-10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-6x+12y=54\\6x-4y=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8y=44\\3x-2y=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{11}{2}\\3x=-5+2y=-5+11=6\end{matrix}\right.\)
=>x=3 và y=11/2
Câu 4:
a: \(\left\{{}\begin{matrix}\left(3-\sqrt{5}\right)x-3y=3+5\sqrt{5}\\4x+y=4-2\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(3-\sqrt{5}\right)x-3y=3+5\sqrt{5}\\12x+3y=12-6\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(15-\sqrt{5}\right)x=3+5\sqrt{5}+12-6\sqrt{5}=15-\sqrt{5}\\12x+3y=12-6\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\3y+12=12-6\sqrt{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\3y=-6\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=-2\sqrt{5}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)x-y=\sqrt{3}\\x+\left(\sqrt{3}+1\right)y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(3-1\right)x-\left(\sqrt{3}+1\right)y=\sqrt{3}\left(\sqrt{3}+1\right)=3+\sqrt{3}\\x+\left(\sqrt{3}+1\right)y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-\left(\sqrt{3}+1\right)y=3+\sqrt{3}\\x+\left(\sqrt{3}+1\right)y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=3+\sqrt{3}+1=4+\sqrt{3}\\x+\left(\sqrt{3}+1\right)y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{3}\\\left(\sqrt{3}+1\right)y=1-x=1-\dfrac{4+\sqrt{3}}{3}=\dfrac{-\sqrt{3}-1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
