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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. 2xy - x2 - y2 + 16
=(2xy-x2-y2)+16
=16-(x-y)2
=(4+x-y)(4-x+y)
b. x2 + x - 6
=x2+3x-2x-6
=x(x+3)-2(x+3)
=(x-2)(x+3)
c. x2 + 5x + 6
=x2+3x+2x+6
=x(x+3)+2(x+3)
=(x+2)(x+3)
a)\(x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
b)\(x^2-5x+5y-y^2=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
a) (x-y)(x+y)-(x+y)=(x+y)(x-y-1) b) (x^2-y^2)-(5x-5y)=(x-y)(x+y)-5(x+y)=(x+y)(x-y-5) c) (x^3+1)-3x(x+1)=(x+1)(x^2-x+1)-3x(x+1)=(x+1)(x^2-4x+1)
a ) ( x3 -x + 3x3y + 3xy2 + y3 -y) = ( x + y )3 - ( x + y ) = ( x-y )2 ( x - y - 1 )
b) x2 + 5x -6 = x2 + 6x -x - 6 = x( x + 6 ) - ( x + 6 ) = ( x -1 ) ( x + 6 )
c) 16 x - 5x2 - 3 = -5x2 + 15x +x -3 = -5x ( x-3 ) + ( x - 3 ) = ( 1 - 5x ) ( x-3)
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)5x2y-10xy2
=5xy(x-2y)
b,:4x(2y-z)+7y(z-2y)
=4x(2y-z)-7y(2y-z)
=(2y-z)(4x-7y)
c,:y(x-z)+7(z-x)
=y(x-z)-7(x-z)
=(x-z)(y-7)
d)36-12x+x^2
=x2-2.x.6+62
=(x-6)2
e) (x-5)^2-16
=(x-5)2-42
=(x-5-4)(x-5+4)
=(x-9)(x-1)
f) 8x^3+1/27
=(2x)3+(1/3)3
=(2x+1/3)(4x2+2/3.x+1/9)
\(3^2-3\)
bằng 6