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a ) \(y\left(x-1\right)=x^2+2\)
\(\Leftrightarrow x^2+2-y\left(x-1\right)=0\)
\(\Leftrightarrow x^2-1-y\left(x-1\right)+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-y\left(x-1\right)=-3\)
\(\Leftrightarrow\left(x-1\right)\left(x+1-y\right)=-3\)
...
b ) \(3xy-5x-2y=3\)
\(\Leftrightarrow9xy-15x-6y=9\)
\(\Leftrightarrow9xy-15x-6y+10=19\)
\(\Leftrightarrow3y\left(3x-2\right)-5\left(3x-2\right)=19\)
\(\Leftrightarrow\left(3y-5\right)\left(3x-2\right)=19\)
...
c ) \(x^2-10xy-11y^2=13\)
\(\Leftrightarrow x^2-11xy+xy-11y^2=13\)
\(\Leftrightarrow x\left(x-11y\right)+y\left(x-11y\right)=13\)
\(\Leftrightarrow\left(x+y\right)\left(x-11y\right)=13\)
...
d ) \(xy-2=2x+3y\)
\(\Leftrightarrow xy-2-2x-3y=0\)
\(\Leftrightarrow y\left(x-3\right)-2\left(x-3\right)-8=0\)
\(\Leftrightarrow\left(y-2\right)\left(x-3\right)=8\)
...
e ) \(5xy+x+2y=7\)
\(\Leftrightarrow5xy+x+2y-7=0\)
\(\Leftrightarrow5x\left(y+\dfrac{1}{5}\right)+2\left(y+\dfrac{1}{5}\right)-\dfrac{37}{5}=0\)
\(\Leftrightarrow\left(5x+2\right)\left(y+\dfrac{1}{5}\right)=\dfrac{37}{5}\)
\(\Leftrightarrow\left(5x+2\right)\left(5y+1\right)=37\)
...
P/s : Vì bài dài nên việc tìm x , y ( lập bảng ) bạn tự làm nhé
Thanks 
a: \(x^4+2x^3+x^2=x^2\left(x+1\right)^2\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
a)\(9y^3-y\)
\(=y\left(9y^2-1\right)\)
\(=y\left(3y-1\right)\left(3y+1\right)\)
\(9y^3-y=y\left(9y^2-1\right)=y\left(3y+1\right)\left(3y-1\right)\)
\(8y^3-2y\left(1-2y\right)^2=2y\left[\left(2y\right)^2-\left(1-2y\right)^2\right]=2y\left(4y-1\right)\)
\(2x^3-8x^2+8x=2x\left(x^2-4x+4\right)=2x\left(x-2\right)^2\)