Calculation (2 × 5 + 2) (4 × 7 + 2) (2004×2007+2)/(1×4+2)(3×6+2)…… (2005×2008+2)

Calculation (2 × 5 + 2) (4 × 7 + 2) (2004×2007+2)/(1×4+2)(3×6+2)…… (2005×2008+2)

n(n+3)+2=(n+1)(n+2)
So molecule = (3 * 4) * (5 * 6) * *(2005*2006)=2006!/2
Denominator = (2 * 3) (4 * 5) * *(2006*2007)=2007!
So the original formula = 2006! / (2 * 2007!) = 1 / 4014

The calculation process of the 12th power of x plus the 8th power of X

X^12+X^8=X^8(X^4+1)

Simplified evaluation: - one third ab - the second power of one fourth a - the second power of one third 2Ab + A, where a = 2, B = 3

-One third ab - the second power of one fourth a - the second power of one third 2Ab + A
=-The second power of a + the second power of a
=-ab
When a = 2, B = 3
Original formula = - 6