Giúp mình với T.T
Phân tích thành nhân tử:
a) 4(xy+4)^2-16(x+y)^2
b) 4(x^2+y^2) - 8(x-ay) - 4(a^2-1)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) ( 3x2 + 3x + 2)2 - ( 3x2 + 3x - 2)2
=(3x2 + 3x + 2 + 3x2 + 3x - 2) [( 3x2 + 3x + 2) - ( 3x2 + 3x - 2) ]
=(6x2+6x)*4
=24x(x+1)
b ) ( xy+1)2 - ( x+y)2
=( xy+1 + x+y ) [( xy+1) - ( x+y)]
=[x(y+1)+(y+1)] [x(y-1) - (y-1)]
=(x+1)(y+1)(x-1)(y-1)
c ) ( x + y)3 - ( x - y)3
=[( x + y)-( x - y)] [( x + y)2 - ( x + y)( x - y) + ( x - y)2 ]
=2y( x2+2xy+y2 - x2+y2+ x2-2xy +y2 )
=2y(3y2+x2)
d ) 4( x2 - y2 ) - 8(x - ay) - 4(a2 - 1)
=4(-a2+2ay-y2+x2-2x+1)
=4[-(a-y)2+(x-1)2]
=-4(y-x-a+1)(y+x-a-1)
1: =(x+y-3x)(x+y+3x)
=(-2x+y)(4x+y)
2: =(3x-1-4)(3x-1+4)
=(3x+3)(3x-5)
=3(x+1)(3x-5)
3: =(2x)^2-(x^2+1)^2
=-[(x^2+1)^2-(2x)^2]
=-(x^2+1-2x)(x^2+1+2x)
=-(x-1)^2(x+1)^2
4: =(2x+1+x-1)(2x+1-x+1)
=3x(x+2)
5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]
=(2x^2+2)*4x
=8x(x^2+1)
6: =(5x-5y)^2-(4x+4y)^2
=(5x-5y-4x-4y)(5x-5y+4x+4y)
=(x-9y)(9x-y)
7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)
=(x^2+2xy+y^2)(x^2-y^2)
=(x+y)^3*(x-y)
8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)
=[(x-2y)^2-4][(x+2y)^2-36]
=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)
a) \(2x-72x^3=2x\left(1-36x^2\right)=2x\left(1-6x\right)\left(1+6x\right)\)
f) \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
a: =(x-z)(y+8)
b; =x^2-2x-3x+6
=(x-2)(x-3)
c: =x^4+10x^2-x^2-10
=(x^2+10)(x^2-1)
=(x^2+10)(x-1)(x+1)
a:=2(x-2)+y(x-2)
=(x-2)(y+2)
b: \(=\left(x+y\right)^2-4\)
=(x+y+2)(x+y-2)
a: \(4x^2-4x\)
\(=4x\cdot x-4x\cdot1\)
\(=4x\left(x-1\right)\)
b: \(x^2-2xy+y^2-4\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
`a, 9x^2 - 16 = (3x+4)(3x-4)`
`b, 4x^2 - 12xy + 9y^2 = (2x-3y)^2`
`c, t^3-8 = (t-2)(t^2 - 2t + 4)`
`d, 2ax^3y^3 + 2a = 2a(x^3y^3 + 1) = 2a(xy+1)(x^2y^2 - xy + 1)`
a) \(\left(9x^2-16\right)=\left(3x-4\right)\left(3x+4\right)\)
b) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
c) \(t^3-8=\left(t-2\right)\left(t^2+2t+4\right)\)
d) \(2ax^3y^3+2a=2a\left(x^3y^3+1\right)\)