\(A=x^6+x^5\left(x-1\right)-x^4\left(x+1\right)+x^3\left(x-1\right)+x^{20}\left(x+1\right)-x\left(x-1\right)+1\)

\(A=x^6-x^6+x^5-x^5-x^4+x^4-x^3+x^3+x^2-x^2+x+1\)

\(A=\left(x^6-x^6\right)+\left(x^5-x^5\right)+\left(x^4-x^4\right)+\left(x^3-x^3\right)+\left(x^2-x^2\right)+x+1\)

\(A=x+1\) x = 999

=> A = 999 + 1 = 1000

\(A=x^6-x^5\left(x-1\right)-x^4\left(x+1\right)+x^3\left(x-1\right)+x^2\left(x+1\right)-x\left(x-1\right)+1\)

\(A=x^6-\left(x^6-x^5\right)-\left(x^5+x^4\right)+\left(x^4-x^3\right)+\left(x^3+x^2\right)-\left(x^2-x\right)+1\)

\(A=x^6-x^6+x^5-x^5-x^4+x^4-x^3+x^3+x^2-x^2+x+1\)

\(A=x+1\)

Thay \(x=999\)vào A, ta có:

\(A=x+1=999+1=1000\)

Vậy tại \(x=999\)thì \(A=1000\)

\(A=1000\)

hok tốt ~!!!

\(=\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{998}{999}=\frac{2\cdot3\cdot4...998}{3\cdot4\cdot5...999}=\frac{2}{999}\)

Vậy xét là \(\frac{1}{2}+1\)nhé.

a,\(\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{1000}{999}\)

=3x4x5x...x1000/2x3x4x...x999

=1000/2=500

b, c tương tự câu a

)(1/2+1)x(1/3+1)x(1/4+1)x...x(1/999+1) 

b)(1/2-1)x(1/3-1)x(1/4-1)x...x(1/1000-1)

c)3/2² x 8/3² x 15/4² x .... x 99/10²

mình ko biết làm chép lại de thui

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