Phân tích các đa thức sau thành nhân tử tổng hợp :
a, x^2 - 25 + y^2 + 2xy
b, 81x^2 - 6yz - 9y^2 - z^2
c, 3x^2 - 6xy + 3y^2
d, 2x^2 + 2y^2 - x^2 z + z - y^2 z - 2
e, x^2 - 2xy + y^2 - 16
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : x2 - y2 - 2x + 2y
= (x2 - y2) - (2x - 2y)
= (x - y)(x + y) - 2(x - y)
= (x - y)(x + y - 2)
a, x2 - y2 - 2x + 2y
= ( x2 - y2 ) - ( 2x - 2y )
= ( x - y ).( x + y ) - 2.( x - y )
= ( x - y ).( x + y - 2 )
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(6x^3y^2.\left(2-x\right)+9x^2y^2\left(x-2\right)\)
\(=6x^3y^2.\left(2-x\right)-9x^2y^2\left(2-x\right)\)
\(=y^2.\left(2-x\right)\left(6x^3-9x^2\right)\)
\(=3x^2y^2.\left(2-x\right)\left(2x-3\right)\)
b. \(x^2-4x+4y-y^2\)
\(=\left(x^2-y^2\right)-\left(4x-4y\right)\)
\(=\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b)x2+2xy+y2-16=(x+y)2-42=(x+y+4)(x+y-4)
c)3x2+5x-3xy-5y=x(3x+5)-y(3x+5)=(3x+5)(x-y)
d)4x2-6x3y-2x2+8x=2x(2x-3x2y-x+4)
e)x2-4-2xy+y2=(x2-2xy+y2)-4=(x-y)2-22=(x-y-2)(x-y+2)
k)x2-y2-z2-2yz=x2-(y+z)2=(x-y-z)(x+y+z)
m)6xy+5x-5y-3x2-3y2=3(x2-2xy+y2)+5(x-y)=3(x-y)2+5(x-y)=(x-y)(3x-3y+5)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(4x^2\right)^2\left(x-y\right)-\left(x-y\right)\)
\(=\left[\left(4x^2\right)^2-1^2\right]\left(x-y\right)\)
\(=\left(4x^2+1\right)\left(4x^2-1\right)\left(x-y\right)\)
\(=\left(4x^2+1\right)\left(2x+1\right)\left(2x-1\right)\left(x-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+2xy+x+2y\)
\(=x\left(x+1\right)+2y\left(x+1\right)\)
\(=\left(x+1\right)\left(2y+x\right)\)
\(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
a)x2+2xy+x+2y
=(2xy+x2)+(2y+x)
=x(2y+x)+(2y+x)
=(x+1)(2y+x)
b)7x2-7xy-5x+5y
=(5y-7xy)+(7x2-5x)
=y(5-7x)-x(5-7x)
=(5-7x)(y-x)
c)x2-6x+9-9y2
=(x2+3xy-3x)-(3xy+9y2-9y)-(3x+9y-9)
=x(x+3y-3)-3y(x+3y-3)-3(x+3y-3)
=(x-3y-3)(x+3y-3)
d)x3-3x2+3x-1+2(x2-x)
Ta thấy x=1 là nghiệm của đa thức
=>đa thức có 1 hạng tử là x-1
=(x-1)(x2+1)
e) (x+y)(y+z)(z+x)+xyz
đề sai
f)x(y2-z2)+y(z2-x2)
=(xy2+yz2)+(x2y+xz2)
=y(xy+z2)-x(xy+z2)
=(y-x)(xy+z2)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) = x2 (y - x) - 9( y - x) = ( y - x ) ( x2 - 9) = ( y -x) ( x - 3 ) ( x + 3)
b) = x2 ( x - 1 ) - 16 ( x - 1 ) = ( x - 1 ) ( x2 - 16 ) = ( x - 1 ) ( x - 4 ) ( x + 4 )
c) = (9x)2 - ( 9y2 + 6yz + z2 ) = (9x)2 - ( 3y + z)2 = (9x - 3y - z ) ( 9x + 3y + z)
d) = z( x - y) - ( x2 -2xy + y2 ) = z(x - y) - (x - y)2 = (x - y) (z - 1)
e) = (x + 3) (x + 5)
f) = (x - 4) ( x + 3)
g) = (9x)2 + 2.9x.2 + 22 - 36x = (9x + 2)2 - (\(6\sqrt{x}\))2 = \(\left(9x+2+6\sqrt{x}\right)\). \(\left(9x+2-6\sqrt{x}\right)\)
a) x2y - x3 - 9y + 9x = x2(y - x) - 9(y - x) = (y - x)(x2 - 9) = (y - x)(x + 3)(x - 3).
b) x2(x - 1) + 16(1 - x) = (x - 1)(x2 - 16) = (x - 1)(x - 4)(x + 4).
c) 81x2 - 6yz - 9y2 - z2 = (9x)2 - ((3y)2 + 2.3yz + z2) = (9x)2 - (3y + z)2 = (9x + 3y +z)(9x - 3y - z).
d) xz - yz - x2 + 2xy - y2 = z(x - y) - (x - y)2 = (x - y)(z - x + y).
e) x2 + 8x + 15 = (x2 + 3x) + (5x + 15) = x(x + 3) + 5(x + 3) = (x + 3)(x + 5).
f) x2 - x - 12 = (x2 - 4x) + (3x - 12) = x(x - 4) + 3(x - 4) = (x - 4)(x + 3).
g) (Đề sai) 81x4 + 4 = (81x4 + 36x2 + 4) - 36x2 = (9x2 + 2)2 - 36x2 = (9x2 + 2 + 6x)(9x2 + 2 - 6x).
a) Ta có: \(x^2-25+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-25\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
c) Ta có: \(3x^2-6xy+3y^2\)
\(=3\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)^2\)
d) Ta có: \(2x^2+2y^2-x^2z+z-y^2z-2\)
\(=2\left(x^2+y^2-1\right)-z\left(x^2+y^2-1\right)\)
\(=\left(x^2+y^2-1\right)\left(2-z\right)\)
e) Ta có: \(x^2-2xy+y^2-16\)
\(=\left(x-y\right)^2-4^2\)
\(=\left(x-y-4\right)\left(x-y+4\right)\)