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22 tháng 10 2020

2. Đặt \(x-1996=t\)

\(\Rightarrow\left(x-1996\right)^3+\left(x-1997\right)^3-1=t^3+\left(t-1\right)^2-1\)

\(=t^3+t^2-2t+1-1=t^3+t^2-2t=t\left(t^2+t-2\right)\)

\(=t.\left[\left(t^2-t\right)+\left(2t-2\right)\right]=t\left[t\left(t-1\right)+2\left(t-1\right)\right]\)

\(=t\left(t-1\right)\left(t+2\right)=\left(x-1996\right)\left(x-1996-1\right)\left(x-1996+2\right)\)

\(=\left(x-1996\right)\left(x-1997\right)\left(x-1994\right)\)

22 tháng 10 2020

1. Đặt x2 + 4x + 8 = y

bthuc ⇔ y2 + 3xy + 2x2

          = y2 + xy + 2xy + 2x2

          = ( xy + y2 ) + ( 2x2 + 2xy )

          = y( x + y ) + 2x( x + y )

          = ( x + y )( y + 2x )

          = ( x + x2 + 4x + 8 )( x2 + 4x + 8 + 2x )

          = ( x2 + 5x + 8 )( x2 + 6x + 8 )

          = ( x2 + 5x + 8 )( x2 + 2x + 4x + 8 )

          = ( x2 + 5x + 8 )[ x( x + 2 ) + 4( x + 2 ) ]

          = ( x2 + 5x + 8 )( x + 2 )( x + 4 )

2. Đặt t = x - 1996 

bthuc ⇔ t3 + ( t - 1 )2 - 1

           = t3 + t2 - 2t + 1 - 1

           = t3 + t2 - 2t

           = t( t2 + t - 2 )

           = t( t2 - t + 2t - 2 )

           = t( t - 1 )( t + 2 )

           = ( x - 1996 )( x - 1996 - 1 )( x - 1996 + 2 )

           = ( x - 1996 )( x - 1997 )( x - 1994 )

3. 4( x2 + 15x + 59 )( x2 + 18x + 72 ) - 3x2 < bó tay :)) >

12 tháng 12 2018

\(a,3\left(x+4\right)-x^2-4x\)

\(=3\left(x+4\right)-\left(x^2+4x\right)\)

\(=3\left(x+4\right)-x\left(x+4\right)\)

\(=\left(3-x\right)\left(x+4\right)\)

\(a,3\left(x+4\right)-x^2-4x\)

\(=3\left(x+4\right)-\left(x^2+4x\right)\)

\(=3\left(x+4\right)-x\left(x+4\right)\)

\(=\left(3-x\right),\left(x+4\right)\)

b. x4 - x2 - 2x - 1

=x4-(x2+2x+1)

=x4-(x+1)2

=(x2-x-1)(x2+x+1)

d. ( x2 + 3x + 1 ) ( x2 + 3x - 3 ) - 5

Đặt x2+3x=y

=> (y+1)(y-3)-5=y2-2y-8=(y-1)2-9

=(y-4)(y+2)

=(x2+3x-4)(x2+3x+2)=(x-1)(x+4)(x+1)(x+2)

21 tháng 11 2016

ôi mai dê

21 tháng 11 2016

mấy bài này max dễ bn đăng từng phần 1 mk lm cho

4 tháng 8 2017

Mình sửa: Bài 1
2)x2+3x-15

20 tháng 5 2018

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)

28 tháng 7 2019

Bài 2:

a) \(x^2+y^2-9-2xy\)

\(=\left(x^2-2xy+y^2\right)-3^2\)

\(=\left(x-y\right)^2-3^2\)

\(=\left(x-y-3\right)\left(x-y+3\right)\)

b) \(4x^2-5x-9\)

\(=4x^2+4x-9x-9\)

\(=4x\left(x+1\right)-9\left(x+1\right)\)

\(=\left(x+1\right)\left(4x-9\right)\)

28 tháng 7 2019

\(\left(2x-3\right)^2-\left(4x-1\right)\left(x+2\right)=4x^2-12x+9-4x^2-7x+2=-19x+11\)

\(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=9x^2-4-9x^2+6x-1=6x-5\)

\(x^2+y^2-9-2xy=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)

\(4x^2-5x-9=\left(4x-9\right)\left(x+1\right)\)

\(\left(x-3\right)^2-\left(x-1\right)\left(x-2\right)=5\Leftrightarrow x^2-6x+9-x^2+3x-2=5\)

\(\Leftrightarrow-3x=-2\Leftrightarrow x=x=\frac{2}{3}\)

\(3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)

2 tháng 11 2018

\(x^8+x+1\)

\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

29 tháng 1 2020

a)\(2x^3-x^2+5x+3=2x^3-2x^2+x^2+6x-x+3\)

\(=\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\)

\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)

\(=\left(x^2-x+3\right)\left(2x+1\right)\)

b)\(27x^3-27x^2+18x-4=27x^3-18x^2-9x^2+12x+6x-4\)

\(=3x\left(9x^2-6x+4\right)-\left(9x^2-6x+4\right)\)

\(=\left(9x^2-6x+4\right)\left(3x-1\right)\)

c)\(4x^4-32x^2+1=4x^4+4x^2-36x^2+1\)

\(=\left(4x^4+4x^2+1\right)-36x^2\)

\(=\left(2x^2+1\right)^2-\left(6x\right)^2\)

\(=\left(2x^2+1+6x\right)\left(2x^2+1-6x\right)\)

2 tháng 8 2016

a, \(x^3+6x^2+11x+6\)

\(=x^3+3x^2+3x^2+9x+2x+6\)

\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+3x+2\right)\)

\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)

\(=\left(x+3\right)\text{[}x\left(x+1\right)+2\left(x+1\right)\text{]}\)

\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)

b, \(2x^3+3x^2+3x+2\)

\(=2x^3+2x^2+x^2+x+2x+2\)

\(=2x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+1\right)\left(2x^2+x+2\right)\)

c, \(x^3-4x^2-8x+8\)

\(=x^3+2x^2-6x^2-12x+4x+8\)

\(=x^2\left(x+2\right)-6x\left(x+2\right)+4\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-6x+4\right)\)