Question

[22 - Đỗ Nhật Minh - 6A17]

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Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x-6y-15=-2x+7y-7\\-24x+3y-6=18x-2y-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7x-13y=-7+15=8\\-42x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42x-78y=48\\-42x+5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-73y=51\\7x-13y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{53}{71}\\7x=13y+8=13\cdot\dfrac{-53}{71}+8=-\dfrac{121}{71}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{53}{71}\\x=-\dfrac{121}{497}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy\\y^2+y-xy-x=y^2-2y+xy-2x-2xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x-y=x-y\\y-x=-2y-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x=0\\3y=-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)