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d)
x3 + 2x2y+ xy2 - 9x
=x*(x2+2xy+y2 -9)
=x*[ (x+y)2 -32 ]
=x * (x+y-3) * (x+y-3)
a) 4x2 - 5xy + y2 = 4x2 - 4xy - xy + y2 = 4x( x - y ) - y( x - y ) = ( x - y )( 4x - y )
b) x2 - 4xy + 3y2 = x2 - xy - 3xy + 3y2 = x( x - y ) - 3y( x - y ) = ( x - y )( x - 3y )
c) 9x2 + 6xy - 8y2 = 9x2 - 6xy + 12xy - 8y2 = 9x( x - 2/3y ) + 12y( x - 2/3y ) = ( x - 2/3y )( 9x + 12y )
d) 2x2 + 3xy - 5y2 = 2x2 - 2xy + 5xy - 5y2 = 2x( x - y ) + 5y( x - y ) = ( x - y )( 2x + 5y )
e) x2 - 35y2 - 2xy = x2 + 5xy - 7xy - 35y2 = x( x + 5y ) - 7y( x + 5y ) = ( x + 5y )( x - 7y )
f) 2x2 + 10xy + 8y2 = 2( x2 + 5xy + 4y2 ) = 2( x2 + xy + 4xy + 4y2 ) = 2[ x( x + y ) + 4y( x + y ) ] = 2( x + y )( x + 4y )
g) x2 - 10xy + 16y2 = x2 - 2xy - 8xy + 16y2 = x( x - 2y ) - 8y( x - 2y ) = ( x - 2y )( x - 8y )
h) 4x2 + 4xy - 15y2 = 4x2 - 6xy + 10xy - 15y2 = 4x( x - 3/2y ) + 10y( x - 2/3y ) = ( x - 2/3y )( 4x + 10y )
i) -7xy + 3x2 + 2y2 = 3x2 - xy - 6xy + 2y2 = 3x( x - 1/3y ) - 6y( x - 1/3y ) = ( x - 1/3y )( 3x - 6y )
j) 56y2 + 4x2 - 36xy = 4( x2 - 9xy + 14y2 ) = 4( x2 - 2xy - 7xy + 14y2 ) = 4[ x( x - 2y ) - 7y( x - 2y ) ] = 4( x - 2y )( x - 7y )
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
\(x^2-y^2-5x-5y\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
học tốt
a) \(6x^2-x-1\)
\(=6x^2-3x+2x-1\)
\(=3x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\)
a) \(x^2-10x+9\)
\(=x^2-9x-x+9\)
\(=x\left(x-9\right)-\left(x-9\right)\)
\(=\left(x-1\right)\left(x-9\right)\)
b) \(3x^2-10xy+3y^2\)
\(=3x^2-9xy-xy+3y^2\)
\(=3x\left(x-3y\right)-y\left(x-3y\right)\)
\(=\left(3x-y\right)\left(x-3y\right)\)
a) Ta có: \(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
b) Ta có: \(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
c) Ta có: \(2x^2-5xy+3y^2\)
\(=2x^2-2xy-3xy+3y^2\)
\(=2x\left(x-y\right)-3y\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-3y\right)\)
d) Ta có: \(16y^3-2x^3-6x\left(x+1\right)-2\)
\(=16y^3-2x^3-6x^2-6x-2\)
\(=2\left[8y^3-x^3-3x^2-3x-1\right]\)
\(=2\left[\left(2y\right)^3-\left(x^3+3x^2+3x+1\right)\right]\)
\(=2\left[\left(2y\right)^3-\left(x+1\right)^3\right]\)
\(=2\left(2y-x-1\right)\left[\left(2y\right)^2+2y\left(x+1\right)+\left(x+1\right)^2\right]\)
\(=2\left(2y-x-1\right)\left(4y^2+2xy+2y+x^2+2x+1\right)\)