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a) \(C=ax-ay-bx+by+\left(y-x\right)^2\)
\(=a\left(x-y\right)-b\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(a-b+x-y\right)\)
b) \(M=\left(5x-10\right)\left(x^2-1\right)\left(3x-6\right)\left(x^2-2x+1\right)\)
\(=5\left(x-2\right)\left(x-1\right)\left(x+1\right)-3\left(x-2\right)\left(x-1\right)^2\)
\(=\left(x-2\right)\left(x-1\right)\left[5\left(x+1\right)-3\left(x-1\right)\right]\)
\(=\left(x-2\right)\left(x-1\right)\left(5x+5-3x+3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(2x+8\right)\)
\(=2\left(x-2\right)\left(x-1\right)\left(x+4\right)\)
1) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)
\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)
\(=x^3.\left(x+2\right)+4x.\left(x+2\right)-5.\left(x+2\right)=\left(x+2\right)\left(x^3+4x-5\right)\)
\(=\left(x+2\right)\left(x^3-x^2+x^2-x+5x-5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)
Đặt \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)
\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)
Thay a , ta có :
\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)
\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
a,2x2-7x+6=(2x2-4x)-(3x-6)
=2x(x-3)-3(x-2)=(x-2)(2x-3)
b,x2+x-6=(x2+3x)-(2x+6)
=x(x-3)-2(x-3)=(x-3)(x-2)
c,x3+3x2+6x+4=x3+x2+2x2+2x+4x+4
=(x+1)(x2+2x+4)
d,x10+x5+1=(x10-x)+(x5-x2)+(x2+x+1)
=x((x3)3-1)+x2(x3-1)+(x2+x+1)
=x(x3-1)(x6+x3+1)+x2(x-1)(x2+x+1)+(x2+x+1)
=x(x-1)(x2+x+1)+x2(x-1)(x2+x+1)+(x2+x+1)
(x2+x+1)(x2-x+x3-x2+1)
e,(12x2-12xy+3y2)-10x(2x-y)=3(4x2-4xy+y2)-10x(2x-y)
=3(2x-y)2-10x(2x-y)=(2x-y)(6x-3y-10x)=(2x-y)(-4x-3y)
phân tích đa thức thành nhân tử
a,2x^2-7x+6
b,x^2+x-6
c,x^3+3x^2+6x+4
d,x^10+x^5+1
e,(12x^2-12xy+3y^2)-10x(2x-y)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
Ta có : 6x2 - 11x + 3
= 6x2 - 2x - 9x + 3
= (6x2 - 2x) - (9x - 3)
= 2x(3x - 1) - 3(3x - 1)
= (2x - 3)(3x - 1)
6) \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)
7) \(x^3+2x^2+3x=x\left(x^2+2x+3\right)\)
8) \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)
9) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=5x\left(x-2y\right)\left(x-3\right)\)
10) \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(3+5x\right)\)
6) 9x3y2 + 3x2y2 = 3x2y2( 3x + 1 )
7) x3 + 2x2 + 3x = x( x2 + 2x + 3 )
8) 6x2y + 4xy2 + 2xy = 2xy( 3x + 2y + 1 )
9) 5x2( x - 2y ) - 15x( x - 2y ) = 5x( x - 2y )( x - 3 )
10 3( x - y ) - 5x( y - x ) = 3( x - y ) + 5x( x - y ) = ( x - y )( 3 + 5x )
a) x3 -2x2 +5x-4
=x3-x2-x2+x+4x-4
=x2(x-1)-x(x-1)+4(x-1)
=(x2-x+4)(x-1)
b) x3-x2+x+3
=x3+x2-2x2-2x+3x+3
=x2(x+1) -2x(x+1)+3(x+1)
=(x2-2x+3)(x+1)
c) 6x3+x2+x+1
=6x3+ 3x2-2x2-x+2x+1
=6x2(x+\(\frac{1}{2}\)) - 2x(x+\(\frac{1}{2}\)) +2(x+\(\frac{1}{2}\))
=(6x2-2x+2) (x+\(\frac{1}{2}\))
=2( 3x2-x+1) (x+\(\frac{1}{2}\))
d) 4x3 + 6x2+4x+1
= 4x3+2x2+4x2+2x+2x+1
= 4x2(x+\(\frac{1}{2}\))+ 4x(x+\(\frac{1}{2}\))+2(x+\(\frac{1}{2}\))
= 2(2x2 +2x+1)( x+\(\frac{1}{2}\))
e) x6 -9x3+8
Ai giúp mk vs