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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
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a: =-3x^2y*x^2y+3x^2y*2xy
=-3x^4y^2+6x^3y^2
b: =x^3-x^2y+x^2y+y^2=x^3+y^2
c: =x*4x^3-x*5xy+2x*x
=4x^4-5x^2y+2x^2
d: =x^3+x^2y+2x^3+2xy
=3x^3+x^2y+2xy
\(AB=\left(x^3-2x^2y+5xy^2-y^2\right)\left(x+2y\right)\)
\(=x^4+2x^3y-2x^3y-4x^2y^2+5x^2y^2+10xy^3-xy^2-2y^3\)
\(=x^4+x^2y^2+10xy^3-xy^2-2y^3\)
\(C=10xy^3-xy^2-2y^3\)
Vậy \(AB-C=x^4+x^2y^2\)