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a,\(\frac{1}{5}x^2y\left(15xy^2-5y+3xy\right)=3x^3y^3-x^2y^2+\frac{3}{5}x^3y^2\)
b,\(5x^3-5x=5x\left(x^2-1\right)=5x\left(x-1\right)\left(x+1\right)\)
c, \(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(3x-5\right)\left(x-y\right)\)
1) 1/5x2y( 15xy2 - 5y + 3xy ) = 3x3y3 - x2y2 + 3/5x3y2
2) a) 5x3 - 5x = 5x( x2 - 1 ) = 5x( x2 - 12 ) = 5x( x - 1 )( x + 1 )
b) 3x2 + 5y - 3xy - 5x = ( 3x2 - 3xy ) + ( 5y - 5x )
= 3x( x - y ) + 5( y - x )
= 3x( x - y ) + 5[ -( x - y ) ]
= 3x( x - y ) - 5( x - y )
= ( 3x - 5 )( x - y )
a)x2-xy+x-y
=x(x-y)+(x-y)
=(x+1)(x-y)
b)3x2-3xy-5x+5y
=3x(x-y)-5(x-y)
=(3x-5)(x-y)
a ) \(x^2-xy+x-y\).
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+1\right).\)
b ) \(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
\(x^2-xy+x-y\)
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
a, \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right).\)
\(b,3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
1)
a) (x+y)3-(x+y)= (x+y)(x+y-1)
b) xem lại đề câu B nha bạn
2)
a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc=0
(a+b)3+c3-3ab(a+b+c)=0
(a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)=0
(a+b+c)(a2+b2+c2-xy-yz-xz)=0
Suy ra: a3+b3+c3=3abc
1. a) = (x+y)3 -(x+y) =(x+y)((x+y)2 -1)
= (x+y)(x+y+1)(x+y-1)
b) = 5(( x-y)2 - 4z2)
= 5( x-y +2z)(x-y-2z)
2. áp dụng ( a+b+c)3 = .....rồi biến đổi
\(A=5x-5y\)
\(=5\left(x-y\right)\)
\(B=3xy^2+x^2y\)
\(=xy\left(3y+x\right)\)
1/a ) = (x+y)3 -(x+y)
= (x+y)[(x+y)2+1]
c) = 5(x2-xy+y2)-20z2
=5(x-y)2-20z2
= 5 [ (x-y)2- 4z2 ]
=5(x-y-4z)(x-y+4z)
Bài 1:
a) x3-x+3x2y+3xy2+y3-y
=x3+2x2y-x2+xy2-xy+x2y+2xy2-xy+y3-y2+x2+2xy-x+y2-y
=x(x2+2xy-x+y2-y)+y(x2+2xy-x+y2-y)+(x2+2xy-x+y2-y)
=(x2+2xy-x+y2-y)(x+y+1)
=[x(x+y-1)+y(x+y-1)](x+y+1)
=(x+y-1)(x+y)(x+y+1)
c) 5x2-10xy+5y2-20z2
=-5(2xy-y2+4z2-2)
Bài 2:
5x(x-1)=x-1
=>5x2-6x+1=0
=>5x2-x-5x+1
=>x(5x-1)-(5x-1)
=>(x-1)(5x-1)=0
=>x=1 hoặc x=1/5
b) 2(x+5)-x2-5x=0
=>2(x+5)-x(x+5)=0
=>(2-x)(x+5)=0
=>x=2 hoặc x=-5
a) x2 + 2x - 8 = x2 - 2x + 4x - 8 = x(x - 2) + 4(x - 2) = (x - 2)(x - 4)
b) x2 + 5x + 6 = x2 + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3)
c) 4x2 - 12x + 8 = 4x2 - 4x - 8x + 8 = 4x(x - 1) - 8(x - 1) = (x - 1)(4x - 8)
d) 3x2 + 8xy + 5y2 = 3x2 + 3xy + 5xy + 5y2 = 3x(x + y) + 5y(x + y) = (x + y)(3x + 5y)
a) x2 + 2x - 8 = x2 - 2x + 4x - 8 = x( x - 2 ) + 4( x - 2 ) = ( x - 2 )( x + 4 )
b) x2 + 5x + 6 = x2 + 2x + 3x + 6 = x( x + 2 ) + 3( x + 2 ) = ( x + 2 )( x + 3 )
c) 4x2 - 12x + 8 = 4( x2 - 3x + 2 ) = 4( x2 - x - 2x + 2 ) = 4[ x( x - 1 ) - 2( x - 1 ) ] = 4( x - 1 )( x - 2 )
d) 3x2 + 8xy + 5y2 = 3x2 + 3xy + 5xy + 5y2 = 3x( x + y ) + 5y( x + y ) = ( x + y )( 3x + 5y )
b) \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=2x\cdot2y\)
\(=4xy\)
c) \(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)
\(=\left(4x+2\right)\cdot2x\)
\(=4x\left(2x+1\right)\)
phần a mình nghĩ là đề sai, sửa thế này không biết có đúng không:
a) \(x\left(x+1\right)-5x-5\)
\(=x\left(x+1\right)-5\left(x+1\right)\)
\(=\left(x-5\right)\left(x+1\right)\)
a,\(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5y\right)\)
b, \(x^2+4x-45\)
\(=x^2+9x-5x-45\)
\(=x\left(x-5\right)+9\left(x-5\right)\)
\(=\left(x-5\right)\left(x+9\right)\)
3x 2 -3xy - 5x + 5y
=(3x2-3xy)-(5x-5y)
=3x(x-y)-5(x-y)
=(x-y)(3x-5)
b) x 2 + 4x – 45
=x2+9x-5x-45
=(x2+9x)-(5x+45)
=x(x-9)-5(x-9)
=(x-9)(x-5)
bài này sai nha kết quả phải bằng (x-5)(x+9) còn bạn đấy là trừ