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a) \(-y^2+\dfrac{1}{9}\)
\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)
\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)
b) \(4^4-256\)
\(=4^4-4^4\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\)\(x^2-y^2-2x+2y\)
\(=\left(x^2-2x+1\right)-\left(y^2-2y+1\right)\)
\(=\left(x-1\right)^2-\left(y-1\right)^2\)
\(=\left(x-y\right)\left(x+y-2\right)\)
\(b)\)\(x^2+4y^2-25+4xy\)
\(=\left(x^2+4xy+4y^2\right)-25\)
\(=\left(x+2y\right)^2-25\)
\(=\left(x+2y-5\right)\left(x+2y+5\right)\)
Dumflinz
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x-y\right)^2-2\left(x-y\right)=\left(x-y\right)\left(x-y-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, x(x-y)+2(x-y)=(x-y)(x+2)
b, \(x^2-6xy+9y^2=\left(x-3y\right)^2\)Thay x=16, y=2 có
\(x^2-6xy+9y^2=\left(x-3y\right)^2=\left(16-2\cdot3\right)^2=10^2=100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(5x^2y+10xy=5xy\left(x+2\right)\)
b, \(x^2-2xy+y^2-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
c, \(x^3-8+2x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\)
\(=\left(x-2\right)\left[\left(x^2+2x+4\right)+2x\right]=\left(x-2\right)\left(x+2\right)^2\)
d, \(x^4+x^2y^2+y^4\):<
a, 5xy( x + 2)
b, ( x - y -5 )(x-y+5)
c,( x-2)( x+ 2)2