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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)
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
1. x2 + 2xy + y2 - xz - yz
= ( x2 +2xy + y2 ) - z ( x + y )
= ( x + y )2 - z ( x + y )
= ( x + y ) [( x + y ) - z ]
= ( x + y ) ( x + y - z )
1 x^2+2xy+y^2-xz-yz
=(x+y)^2-z(x+y)
=(x+y)(x+y-z)
2 (7x^2-14xy+7^2)-29z^2
=7(x^2-2xy+1)-29z^2
=7(x-1)^2-29z^2
=7(x-1)^2-25z^2-7z^2
=7(x-1-5)(x-1+5)-7z^2
=7(x-6)(x+4)-7z^2
=7((x+6)(x+4)-z^2)
3 5x^3-5x^2y+10x^2-10xy
=5x(x^2-xy+2x-2y)
4 5x^2-10xy+5y^2-20z^2
=5(x^2-2xy+y^2)-20z^2
=5(x+y)^2-20z^2
=5((x+y)^2-4z^2)
=5((x+y-2z)(x+y+2z))
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. 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)
\(5x^2-10xy^2+5y^4\)
\(=5.\left(x^2-2xy^2+y^4\right)\)
\(=5.\left[x^2-2xy^2+\left(y^2\right)^2\right]\)
\(=5.\left(x-y^2\right)^2\)
a) x4 + 2x3 + x2 = x2.(x2 + 2x + 1) = x2(x + 1)2
b) x3 - x + 3x2y + 3xy2 + y3 - y = x3 + 3x2y + 3xy2 + y3 - x - y = (x + y)3 - (x + y) = (x + y)[(x + y)2 - 1] = (x + y - 1)(x + y)(x + y + 1)
c) 5x2 - 10xy + 5y2 - 20z2 = 5.(x2 - 2xy + y2 - 4z2) = 5[(x - y)2 - (2z)2] = 5(x - y - 2z)(x - y + 2z)
\(a,x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
\(b,x^3-x+3x^2y+3xy^2+y^3-y=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(c,5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left[\left(x-y+2z\right)\left(x-y-2z\right)\right]\)
\(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
5(x2-2xy+y2-4-4z2)