Ta có: \(x^4+64\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot8+64-2\cdot x^2\cdot8\)

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

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

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

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

\(=\left(x+2\right)\left[3\left(x-2\right)-1\right]=\left(x+2\right)\left(3x-7\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-8\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-8\)

\(=\left(x^2+7x+11\right)^2-9\)

\(=\left(x^2+7x+11-3\right)\left(x^2+7x+11+3\right)=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

x⁴ + 1024 = x⁴ + 64x² + 1024 - 64x²

= (x² + 32)² - (8x)²

= (x² - 8x + 32)(x² + 8x + 32)

(x+2)(x+3)(x+4)(x+5)-24

= [(x+2)(x+5)][(x+3)(x+4)] -24

=(x^2+7x+10)(x^2+7x+12)-24

thay x^2+7x+11=y

=> (y-1)(y+1)-24=y^2-1^2-24=y^2-25=(y-5)(y+5)

= (x^2+7x+11-5)(x^2+7x+11+5)=(x^2+7x+6)(x^2+7x+16)=(x^2+x+6x+6)(x^2+7x+16)=[x(x+1)+6(x+1)]((x^2+7x+16)=(x+1)(x+6)(x^2+7x+16)

(x + 2)(x + 3)(x + 5)(x + 7) - 24

= [(x + 2)(x + 5)][(x + 3)(x + 4)] - 24

=(x² + 7x + 10)(x² + 7x +12) - 24

Đặt x² + 7x + 11 = t ; ta có:

(t - 1)(t + 1) - 24

= t² - 1² - 24

= t² - 25

= (t - 5)(t + 5)

Thay t = x² + 7x + 11 ta được:

(x² + 7x + 11 - 5)(x² + 7x +11 + 5)

= (x² + 7x + 6)(x² + 7x + 16)

= (x + 1)(x + 6)(x² + 7x + 16)

Chúc bn học tốt

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)( * )

Đặt \(t=x^2+7x+10\), khi đó (*) trở thành:

\(t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)

Thay \(t=x^2+7x+10\) vào, ta được:

\(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

Vậy ...

(x + 1)(x + 2)(x + 3)(x + 4) - 24

= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 24

= (x² + 4x + x +4)(x² + 3x + 2x + 12) - 24

= (x² + 5x + 4)(x² + 5x + 12) - 24

Đặt t = x² + 5x + 8

Ta có: x² + 5x + 4 = x² + 5x + 8 - 4 (1)

x² + 5x + 12 = x² + 5x + 8 + 4 (2)

Thay t = x² + 5x + 8 vào (1) và (2), ta có:

⇒ (t - 4)(t + 4) - 24

= t² - 16 - 24

= t² - 40

= (t - \(\sqrt{40}\))(t + \(\sqrt{40}\))

= (x² + 5x + 8 - \(\sqrt{40}\))(x² + 5x + 8 + \(\sqrt{40}\))

\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)

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

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

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

\(=\left(x^2+5x+5\right)^2\)

\(x^5+x^4+1\)

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

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

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

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