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PT
0
PT
0
PQ
4
25 tháng 1 2017
6x^5+15x^4+20x^3+15x^2+6x+1
=3x^4(2x+1)+6x^3(2x+1)+7x^2(2x+1)+4x(2x+1)+(2x+1)
=(2x+1)(3x^4+6x^3+7x^2+4x+1)
=(2x+1)(3x^2(x^2+x+1)+3x(x^2+x+1)+(x^2+x+1)
=(2x+1)(x^2+x+1)(3x^2+3x+1)
DV
1
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 3 2019
a/ \(\left(x^2-x+2\right)^2+\left(x-2\right)^2=\left(x^2-x+2\right)^2-x^2+x^2+\left(x-2\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2\right)+2x^2-4x+4\)
\(=\left(x^2-2x+2\right)\left(x^2+2\right)+2\left(x^2-2x+2\right)\)
\(=\left(x^2-2x+2\right)\left(x^2+4\right)\)
b/ \(6x^5+15x^4+20x^3+15x^2+6x+1\)
\(=6x^5+6x^4+2x^3+9x^4+9x^3+3x^2+9x^3+9x^2+3x+3x^2+3x+1\)
\(=2x^3\left(3x^2+3x+1\right)+3x^2\left(3x^2+3x+1\right)+3x\left(3x^2+3x+1\right)+3x^2+3x+1\)
\(=\left(3x^2+3x+1\right)\left(2x^3+3x^2+3x+1\right)\)
\(=\left(3x^2+3x+1\right)\left(x^3+\left(x+1\right)^3\right)\)
\(=\left(3x^2+3x+1\right)\left(2x+1\right)\left(x^2-\left(x+1\right)x+\left(x+1\right)^2\right)\)
\(=\left(3x^2+3x+1\right)\left(3x+1\right)\left(x^2+x+1\right)\)
29 tháng 1 2020
a) \(6x^2-x-1\)
\(=6x^2-3x+2x-1\)
\(=3x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\)
QN
1
23 tháng 10 2015
a)x5+x2-x2+x-1 = x2(x3+1) - (x2-x+1)
=x2(x+1)(x2-x+1) - (x2-x+1)
=(x2-x+1)(x3+x2-1)
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)
a)\(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
\(=x^4-2x^3+6x^2-8x+8\)
\(=x^4-2x^3+2x^2+4x^2-8x+8\)
\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)\)
\(=\left(x^2+4\right)\left(x^2-2x+2\right)\)
b)\(6x^5+15x^4+20x^3+15x^2+6x+1\)
\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)
\(=3x^4\left(2x+1\right)+6x^3\left(2x+1\right)+7x^2\left(2x+1\right)+4x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(3x^4+6x^3+7x^2+4x+1\right)\left(2x+1\right)\)
\(=\left[3x^4+3x^3+x^2+3x^3+3x^2+x+3x^2+3x+1\right]\left(2x+1\right)\)
\(=\left[x^2\left(3x^2+3x+1\right)+x\left(3x^2+3x+1\right)+\left(3x^2+3x+1\right)\right]\left(2x+1\right)\)
\(=\left(x^2+x+1\right)\left(3x^2+3x+1\right)\left(2x+1\right)\)