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a: \(=\left(x+2-y\right)\left(x+2+y\right)\)
c: \(=\left(x-y\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(=3x\left(x+2y\right)\)
c)\(=\left(x-7\right)\left(x-1\right)\)
b)\(=x\left(x-2y\right)+3\left(x-2y\right)=\left(x+3\right)\left(x-2y\right)\)
d)\(=\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
\(a,3x^2+6xy=3x\left(x+2y\right)\\ c,x^2-8x+7=\left(x^2-x\right)-\left(7x-7\right)=x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(x-7\right)\\ b,x^2-2xy+3x-6y=\left(x^2+3x\right)-\left(2xy+6y\right)=x\left(x+3\right)-2y\left(x+3\right)=\left(x+3\right)\left(x-2y\right)\\ d,4x^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
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a: x^2+4xy-21y^2
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b: \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
=5x(x+y)+y(x+y)
=(x+y)(5x+y)
c: \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
=x(x+5y)-3y(x+5y)
=(x+5y)(x-3y)
d: \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
=x(x-2y)-5y(x-2y)
=(x-2y)(x-5y)
a) \(x^2+4xy-21y^2\)
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b) \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(5x+y\right)\left(x+y\right)\)
c) \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
\(=x\left(x+5y\right)-3y\left(x+5y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
d) \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-5y\right)\left(x-2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3x^2-6x+9x^2=12x^2-6x=6x\left(2x-1\right)\\ b,3x^2+5y-3xy-5x=\left(3x^2-3xy\right)-\left(5x-5y\right)=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\\ c,3y^2-3z^2+3x^2+6xyz=3\left(y^2-z^2+x^2+2xyz\right)\\ d,x^2-25-2xy+y^2=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(x^2y^2-x^2+6xy-9y^2\)
\(=x^2y^2-\left(x^2-6xy+y^2\right)\)
\(=\left(xy\right)^2-\left(x-3y\right)^2\)
\(=\left(xy-x+3y\right)\left(xy+x-3y\right)\)
b) Ta có: \(9-x^2+2xy-y^2\)
\(=9-\left(x^2-2xy+y^2\right)\)
\(=9-\left(x-y\right)^2\)
\(=\left(9-x+y\right)\left(9+x-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2x^3y-2xy=2xy\left(x^2-1\right)=2xy\left(x-1\right)\left(x+1\right)\)
\(b,x^2-2x-4x^2-4x=-3x^2-2x-4x\\ =-3x^2-6x=-3\left(x^2+2x\right)=-3x\left(x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
sai rồi vì (x2+6xy+9) ko có y2 nên ko thể có hằng đẳng thức được
\(2xy^2-6xy^2=-4xy^2\)