2: Ta có: 2(x+1)(2y+1)+x+2=0
=>\(2\left(2xy+x+2y+1\right)+x+2=0\)
=>4xy+2x+4y+2+x+2=0
=>4xy+4y+4+3x=0
=4y(x+1)+3x+3=-1
=>(x+1)(4y+3)+1=0
Nếu \(y=-\frac34\) thì \(\left(x+1\right)\cdot0+1=0\)
=>1=0(vô lý)
=>y<>-3/4
=>\(\left(x+1\right)\left(4y+3\right)=-1\)
=>\(x+1=-\frac{1}{4y+3}\)
=>\(x=-\frac{1}{4y+3}-1=\frac{-1-4y-3}{4y+3}=\frac{-4y-4}{4y+3}\)
Ta có: \(4xy^2+4y^2+4xy-x-2y-6=0\)
=>\(4y^2\left(x+1\right)+4xy-x-2y-6=0\)
=>\(4y^2\cdot\frac{-1}{4y+3}+4y\cdot\frac{-4y-4}{4y+3}+\frac{4y+4}{4y+3}-2y-6=0\)
=>\(\frac{-4y^2-16y^2-16y+4y+4+\left(-2y-6\right)\left(4y+3\right)}{4y+3}=0\)
=>\(-20y^2-12y+4-8y^2-6y-24y-18=0\)
=>\(-28y^2-42y-14=0\)
=>\(2y^2+3y+1=0\)
=>(y+1)(2y+1)=0
=>\(\left[\begin{array}{l}y=-1\left(nhận\right)\\ y=-\frac12\left(nhận\right)\end{array}\right.\)
Khi y=-1 thì \(x=\frac{-4\cdot\left(-1\right)-4}{4\cdot\left(-1\right)+3}=\frac{4-4}{-4+3}=0\)
Khi y=-1/2 thì \(x=\frac{-4\cdot\frac{-1}{2}-4}{4\cdot\frac{-1}{2}+3}=\frac{2-4}{-2+3}=\frac{-2}{1}=-2\)
