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Câu 1 : x2-y2+2yz-z2=-(y2-2yz+z2-x2) Câu 2: x2-2xy+y2-xz+yz=(x2-2xy+y2)-xz+yz
=-(y-z)2 -x2 =(x-y)2-z(x-y)
=-(y-z-x)(y-z+x) =(x-y)(x-y-z)
![](https://rs.olm.vn/images/avt/0.png?1311)
k) = x( 2x - 1 ) - 3y( 2x - 1 ) = ( 2x - 1 )( x - 3y )
l) = x( x - y ) + 5( x - y ) = ( x - y )( x + 5 )
m) = ( a2 - 4a + 4 )( a2 + 4a + 4 ) = ( a - 2 )2( a + 2 )2
n) = y2( x2 - 1 ) - ( x2 - 1 ) = ( x - 1 )( x + 1 )( y - 1 )( y + 1 )
q) = 3[ ( x - y )2 - 4z2 ] = 3( x - y - 2z )( x - y + 2z )
![](https://rs.olm.vn/images/avt/0.png?1311)
x2 - 2xy - 4z2 + y2
= (x2 - 2xy+y2) - (2z)2
= (x-y)2- (2z)2
= (x-y-2z)(x-y+2z)
x2-2xy-4z2+y2
=(x2-2xy+y2)-(2z)2
=(x-y)2-(2z)2
=(x-y-2z)(x-y+2z)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x3 - 3x2 - 4x + 12
= x2(x-3) - 4(x-3)
= (x-3)(x2 -4)
=(x-3)(x-2)(x+2)
\(x^3-3x^2-4x+12\)
\(=\left(x^3-3x^2\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5x2 - 10xy + 5y2 - 20z2
=5(x2 - 2xy + y2 - 4z2)
= 5[ (x2 - 2xy + y2) - (2z)2]
= 5[(x-y)2 - (2z)2]
= 5(x-y-2z)(x-y+2z)
5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[(x-y)2-(2z)2]
=5(x-y-2z)(x-y+2z)
Câu 1: 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2 - 4z2)
= 3[(x2 -2xy+y2)- (2z)2]
= 3[(x-y)2 - (2z)2]
=3(x-y-2z)(x-y +2z)
Câu 2: x2 - 6xy - 25z2+ 9y2
= x2 - 2x.3y + (3y)2 - (5z)2
= (x-3y)2 - (5z)2
= (x-3y-5z)(x-3y+5z)