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CW
15 tháng 7 2016
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
NP
1
23 tháng 7 2018
\(a,-2x^3y^4-4x^4y^3+2x^3y^3\)
\(=-2x^3y^3\left(y+x-1\right)\)
\(b,ab\left(x-5\right)-a^2\left(5-x\right)\)
\(=\left(x-5\right)\left(ab+a^2\right)\)
SM
19 tháng 9 2018
A = 6x4 - 5x3 + 4x2 + 2x - 1
= 6x4 + 3x3 - 8x3 - 4x2 + 8x2 + 4x - 2x - 1
= 3x3. ( 2x + 1 ) - 4x2 ( 2x + 1 ) + 4x ( 2x + 1 ) - ( 2x + 1 )
= ( 2x + 1 ) ( 3x3 - 4x2 + 4x - 1 )
= ( 2x + 1 ) ( 3x3 - x2 - 3x2 + x + 3x - 1 )
= ( 2x + 1 ) [ x2 ( 3x - 1 ) - x ( 3x - 1 ) + ( 3x - 1 ) ]
= ( 2x + 1 ) ( 3x - 1 ) ( x2 - x + 1 )
B = 4x4 + 4x3 + 5x2 + 8x - 6
= 4x4 - 2x3 + 6x3 - 3x2 + 8x2 - 4x + 12x - 6
= 2x3 ( 2x - 1 ) + 3x2 ( 2x - 1 ) + 4x ( 2x - 1 ) + 6 ( 2x - 1 )
= ( 2x - 1 ) ( 2x3 + 3x2 + 4x + 6 )
= ( 2x - 1 ) [ x2 ( 2x + 3 ) + 2 ( 2x + 3 ) ]
= ( 2x - 1 ) ( 2x + 3 ) ( x2 + 2 )
C = x4 + x3 - 5x2 + x - 6
= x4 - 2x3 + 3x3 - 6x2 + x2 - 2x + 3x - 6
= x3 ( x - 2 ) + 3x2 ( x - 2 ) + x ( x - 2 ) + 3 ( x - 2 )
= ( x - 2 ) ( x3 + 3x2 + x + 3 )
= ( x - 2 ) [ x2 ( x + 3 ) + ( x + 3 ) ]
= ( x - 2 ) ( x + 3 ) ( x2 + 1 )
9 tháng 10 2018
a, 4x2 - 12x + 9
= (2x + 3)2
b, 9x4y3 + 3x2y4
= 3x2y3(3x2 + y)
c, ( x - 3 )2 - 2x ( x - 3 )
= (x - 3)(x - 3 - 2x)
= (x - 3)(-x - 3)
d, 3x ( x - 1 ) + 6 ( x - 1 )
= 3(x - 1)(x + 2)
e, 2x ( x + 1 ) - 4x - 4
= 2x(x + 1) - 4(x + 1)
= (x + 1)(2x - 4)
= 2(x + 1)(x - 2)
f, ( 2x - 3 )2 - 4x + 6
= (2x - 3)2 - 2(2x - 3)
= (2x - 3)(2x - 3 - 2)
= (2x - 3)(2x - 5)
28 tháng 1 2020
1) \(x^2+2xy+y^2-x-y-12\)
= \(\left(x+y\right)^2-\left(x+y\right)-12\)
Đặt \(x+y=z\) (đặt ẩn phụ)
\(\Rightarrow z^2-z-12\)
\(=z^2+3z-4z-12\)
\(=z\left(z+3\right)-4\left(z+3\right)\)
\(=\left(z+3\right)\left(z-4\right)\)
Khi đó: \(\left(x+y+3\right)\left(x+y-4\right)\)
#HuyenAnh
2 tháng 8 2018
\(x^4+4x^2-5\)
\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-9\)
\(=\left(x^2+2\right)^2-9\)
\(=\left(x^2+2+3\right)\left(x^2+2-3\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)
a) \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)
\(=x^2\left(x+1\right)\left[x^2\left(x-1\right)+2\right]\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
b) Ta có: \(4x^4+y^4\)
\(=4x^4+y^4+4x^2y^2-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2\)
\(=\left(2x^2-2xy+y^2\right)\left(2x^2+2xy+y^2\right)\)
a, \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)
\(=x^2\left[x^2\left(x-1\right)\left(x+1\right)+2\left(x+1\right)\right]=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
b, \(4x^4+y^4=\left(2x^2\right)^2+2.2x^2.y^2+\left(y^2\right)^2-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+y^2-2xy\right)\left(2x^2+y^2+2xy\right)\)