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a )
\(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-4z^2\right]\)
b )
\(-5x^2-16x-3\)
\(=-5x^2-15x-x-3\)
\(=-5x\left(x+3\right)-\left(x+3\right)\)
\(=\left(-5x-1\right)\left(x+3\right)\)
c )
\(x^2-5x+5y-y^2\)
\(=\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)-5\right]\)
d )
\(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-4z^2\right]\)
P/s : Mình bổ sung :
a )
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
d )
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
ta có :\(5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left(\left(x-y\right)^2-\left(2z\right)^2\right)=5\left(x-y-2z\right)\left(x-y+2z\right)\)
d)
x3 + 2x2y+ xy2 - 9x
=x*(x2+2xy+y2 -9)
=x*[ (x+y)2 -32 ]
=x * (x+y-3) * (x+y-3)
5x\(^2\)- 10xy +5x\(^2\)-20z\(^2\)
= 5(x\(^2\)-2xy+x\(^2\)-4z\(^2\))
= 5(2x\(^2\)-2xy-4z\(^2\))
5^2-10xy+5x^2-20z^2
=5(x^2-2xy+y^2-4z^2)
=5((x-y)^2-4z^2)
=5(x-y-2z)(x-y+2z)
a) (x^2+x)^2-14(x^2+x)+24
=(x^2+x)^2-2(x^2+x)-12(x^2+x)24
=(x^2+x)(x^2+x-2)-12(x^2+x-2)
=(x^2+x-12)(x^2+x-2)
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
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, = (x^2+10x+25)-y62 = (x+5)^2-y^2 = (x+5-y).(x+5+y)
b, = xy.(x-y)
c, = (x-y).(x+y)+5.(x-y) = (x-y).(x+y+5)
k mk nha
a. 5(x^2-2xy+y^2-4z^2)=5[(x-1)^2-(2z)^2]=5(x-1-2z)(x-1+2z)
b.6x^2-23x-18=6^2-4x+27x-18= 2x(3x-2)+9(3x-2)=(2x+9)(3x-2)