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21 tháng 7 2016

https://coccoc.com/search/math#query=Ph%C3%A2n+t%C3%ADch+%C4%91a+th%E1%BB%A9c+th%C3%A0nh+nh%C3%A2n+t%E1%BB%AD%3A+x%5E8%2B98x%5E4%2B1

21 tháng 7 2016

x8 + 98x4 + 1 = (x8 + 2x4 + 1 ) + 96x4

= (x4 + 1)2 + 16x2(x4 + 1) + 64x4 - 16x2(x4 + 1) + 32x4

= (x4 + 1 + 8x2)2  – 16x2(x4 + 1 – 2x2) = (x4 + 8x2  + 1)2  - 16x2(x2 – 1)2

= (x4 + 8x2  + 1)2  - (4x3 – 4x )2

= (x4 + 4x3 + 8x2  – 4x + 1)(x4 - 4x3 + 8x2  + 4x + 1)

Ta có : \(x^8+14x^4+1\)

\(=x^8+2.x^4.7+1\)

\(=x^8+2.x^4.7+49-48\)

\(=\left(x^4+7\right)^2-48\)

\(=\left(x^4+7-\sqrt{48}\right)\left(x^4+7+\sqrt{48}\right)\)

3 tháng 6 2018

a/\(=\left(x^4+1\right)^2+12x^4=\left(x^4+1\right)^2+4x^2\left(x^4+1\right)+4x^4-4x^2\left(x^4+1\right)+8x^4\)

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

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

b/\(=\left(x^4+1\right)^2+96x^4=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)

\(=\left(x^4+1+8x^2\right)^2-16x^2\left(x^4+1-2x^2\right)=\left(x^4+8x^2+1\right)-\left(4x^3-4x\right)^2\)

\(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)

16 tháng 11 2018

a, \(x^3-x^2-4\)

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

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

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

16 tháng 11 2018

a) \(x^3-x^2-4\)

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

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

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

b) \(x^8-98x^4+1\)

\(=\left(x^4\right)^2+2\cdot x^4\cdot1+1^2-100x^4\)

\(=\left(x^4+1\right)^2-\left(10x^2\right)^2\)

\(=\left(x^4-10x^2+1\right)\left(x^4+10x^2+1\right)\)

\(\left(a\right)x^8+98x^4+1\)

\(\text{ Phân tích thành nhân tử}\)

\(\left(x^4-4x^3+8x^2+4x+1\right)\left(x^4+4x^3+8x^2+\left(-4\right)x+1\right)\)

\(\left(b\right)4x^4-32x^2+1\)

\(\text{ Phân tích thành nhân tử}\)

\(-\left(28x^2-1\right)\)

1 tháng 10 2017

cái này phân tích thành nhân tử:

vì máy tính nên ko viết đc mũ

(x mũ 4-4xmũ 3+8x mũ 2+4x+1)vì vậy biểu thức ko thể rút gọn

b) x7 + x2 + 1 = (x7 – x) + (x2 + x + 1) 
= x.(x6 – 1) + (x2 + x +1) 
= x.(x3 - 1).(x3 +1) + (x2 + x +1) 
= x.(x-1).(x2 + x +1).(x3 +1) + (x2 + x +1) 
= (x2 + x +1).[x.(x-1).(x3 +1) + 1] 
= (x2 + x +1).[(x2-x).(x3 +1) + 1] 
= (x2 + x +1).(x5-x4 + x2 -x + 1

2 tháng 9 2017

\(h\left(x\right)=x^7+x^5+1=x^7+x^6+x^5-x^6+1=x^5\left(x^2+x+1\right)-\left(x^3+1\right)\left(x^3-1\right)\)

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

3 tháng 11 2018

a) x4 + 1997x2 + 1996x +1997

= x4 + 1997x2 + 1997x - x +1997

=(x4-x) + (1997x2 +1997x+1997)

=x(x3-1) + 1997(x2+x+1)

=x(x-1)(x2+x+1) + 1997(x2+x+1)

=(x2+x+1)(x2-x) + 1997(x2+x+1)

=(x2+x+1)(x2-x+1997)

b) x2 -x -2001.2002

=x2 - x -2002+2002

=(x2-20022)-(x-2002)

=(x-2002)(x+2002) - (x-2002)

=(x-2002)(x+2002+1)

=(x-2002)(x+2003)

c)x8 + 98x4 +1

= (x8+2x4+1) + 96x4

= (x4+1)2 + 96x4

=[(x4+1)2 + 2.(x4+1).8 + 64x4 ]+[32x4 - 16x2(x4+1)]

=(x4+1+8x2)-16x2(-2x2+x4+1)

=(x4+8x2+1)2- 16x2(x2-1)2

=(x4 + 8x2 +1)2- [4x(x2-1)]2

=(x4+8x2+1)2 - (4x3-4x)2

=(x4-4x3+8x2+4x+1)(x4+4x3+8x2-4x+1)

16 tháng 6 2017

a)\(3x^2-8x+4\)

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

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

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

b)\(4x^4+81\)

\(=4x^4+36x^2+81-36x^2\)

\(=\left(2x^2+9\right)^2-36x^2\)

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

c)\(x^8+98x^4+1\)

\(=\left(x^8+2x^4+1\right)+96x^4\)

\(=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)

\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)

\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)

\(=\left(x^4+8x^2+1\right)^2-\left(4x^3-4x\right)^2\)

\(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)

d)\(x^4+6x^3+7x^2-6x+1\)

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

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

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