a, \(x^3-x^2-4\)

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

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

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

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) \(x^8-98x^4+1\)

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

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

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

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

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

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

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

b) 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

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

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

Ta có : \(x^8+14x^4+1\)

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

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

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

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

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

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

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

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

\(\text{ Phân tích thành nhân tử}\)

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

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

\(\text{ Phân tích thành nhân tử}\)

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

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

Bạn kia làm đúng rồi

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

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)

a) \(x^5-x^4-1\)

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

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

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

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

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

b) \(x^8+x^7+1\)

\(=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)

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

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

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

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

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

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

a) \(x^5-x^4-1=x^5+x^2-x^4-x^2-1\)

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

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

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

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

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

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

b) \(x^8+x^7+1=x^8+x^7+x^6-x^6+1\)

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

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

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

Mong cô Chuy cho e thêm 1 Gp nựa nha cô '-'

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 -2002² +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)