Các bạn giúp mk với

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{999}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{998}{999}=\frac{1.2.3.4.....998}{2.3.4.5.....999}=\frac{1}{999}\)

\(\frac{\left(1-1\right)}{2}x\frac{\left(1-1\right)}{3}x\frac{\left(1-1\right)}{4}\frac{\left(1-1\right)}{999}\)

\(=\frac{x^2}{23976}\)

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

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

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

\(A=x+1\) 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 ~!!!

1/1 x 2 + 1/2 x 3 + 1/3 x 4 + ... + 1/999 x 1000 + 1

= 1/1 - 1/1000 + 1 

= 999/1000 + 1

= 1999/1000

Chuc ban may man

1,999 nha

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

tự tính đi nhớ

2, 

Q=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)x\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

  = \(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)x\left(\frac{3+\left(-2\right)+\left(-1\right)}{6}\right)\)

 =\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)x0\)

Vậy Q=0

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{999\times1000}+1\)

\(=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{999000}+1\)

\(=1-\frac{1}{1000}+1\)

\(=\frac{1999}{1000}\)

bằng 1999/1000

bài này mình làm được nhưng mà dài vậy sao làm nổi 

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