Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(x^4-9x^2\)
\(=x^2\left(x^2-9\right)\)
\(=x^2\left(x-3\right)\left(^{ }x+3\right)\)
b) \(3x^2-12x+12\)
\(=3x\left(x^2-4x+4\right)\)
\(=3x\left(x-2\right)^2\)
c) \(x^2+5x+6\)
\(=x^2+3x+2x+6\)
\(=x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x+2\right)\)
x4 - 9x2
= x4 - ( 3x )2
= ( x2 - 3x ) ( x2 + 3x )
b) 3x3 - 12x2 + 12x
= 3x3 - 6x2 - 6x2 + 12x
= 3x2( x - 2 ) - 6x ( x - 2 )
= ( 3x2 - 6x ) ( x - 2 )
= 3x ( x - 2 ) ( x - 2 )
= 3x ( x- 2 )2
c) x2 + 5x + 6
= x2 + 2x + 3x + 6
= x ( x + 2 ) + 3 ( x + 2 )
= ( x + 3 ) ( x + 2 )
a)\(x^3+x+2=x^3+1+x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+2\right)\)
b)\(x^3+3x^2-4=x^3-1+3x^2-3\)
\(=\left(x-1\right)\left(x^2+x+1\right)+3\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left[x^2+x+1+3x+3\right]\)
\(=\left(x-1\right)\left(x+2\right)^2\)
a) A=x3+3x2+3x
A=x3+3x2.1+3x.12+13
A=(x+1)3
b)A=x3-3x2+3x-1
A=x3-3x2.1+3x.12-13
A=(x-1)3
c)A=x3+6x2+12x
A=x3+3.2x2+3.22x+13
A=(x+1)3
a) ( 4x+1) (12x-1) (3x+2) (x+1) -4
=(4x+1)(3x+2)(12x-1)(x+1)-4
=(12x2+11x+2)(12x2+11x-1)-4
Đặt t=12x2+11x+2 ta được:
t.(t-3)-4
=t2-3t-4
=t2+t-4t-4
=t.(t+1)-4.(t+1)
=(t+1)(t-4)
thay t=12x2+11x+2 ta được:
(12x2+11x+3)(12x2+11x-2)
Vậy ( 4x+1) (12x-1) (3x+2) (x+1) -4=(12x2+11x+3)(12x2+11x-2)
b) (x2+2x)2+9x2+18x+20
=(x2+2x)2+9.(x2+2x)+20
Đặt y=x2+2x ta được:
y2+9y+20
=y2+4y+5y+20
=y.(y+4)+5.(y+4)
=(y+4)(y+5)
thay y=x2+2x ta được:
(x2+2x+4)(x2+2x+5)
Vậy (x2+2x)2+9x2+18x+20=(x2+2x+4)(x2+2x+5)
\(^{x^4+3x^3+x^2-12x-20}\)
= x^4 + 2X^3 + x^3 + 2x^2 - X^2 - 2X -10X - 20
= X^3( x + 2 ) + X^2( x+2) -x(x+2) - 10(x+2)
= ( x^3 + x^2 - x -10) (x+2 )
= (x-2)( x^2 + 3x+5)(x+2)
x^4+3x^2+x^2-12x-20
=x^4+2x^3+x^3+2x^2-x^2-2x-10x-20
=x^3(x+2)+x^2(x+2)-x(x+2)-10(x+2)
=(x+2)(x^3+x^2-x-10)
=(x+2)(x^3-2x^2+3x^2-6x+5x-10)
=(x+2)[x^2(x-2)+3x(x-2)+5(x-2)]
=(x+2)(x-2)(x^2+3x+5)