Ta đặt biểu thức đã cho là A

suy ra A < (10-1)/10! + (11-1)/11! +...+ (1000-1)/1000!

=> A < 10/10! - 1/10! + 11/11! - 1/11! +...+ 1000/1000! - 1/1000!

=> A < 1/9! - 1/10! + 1/10! - 1/11! +...+ 1/999! - 1/1000!

=> A < 1/9! - 1/1000! < 1/9!

Vậy A < 1/9!

Chúc bạn hoc tốt

Bạn tham khảo nhé 

\(a)\)Đặt  \(A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\)

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A< 1-\frac{1}{100}=\frac{100-1}{100}=\frac{99}{100}< 1\) ( đpcm ) 

Vậy \(A< 1\)

\(\frac{9}{10!}+\frac{10}{11!}+...+\frac{999}{1000!}\)

\(=\frac{10-1}{10!}+\frac{11-1}{11!}+...+\frac{1000-1}{1000!}\)

\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+...+\frac{1}{999!}-\frac{1}{1000!}\)

\(=\frac{1}{9!}-\frac{1}{1000!}< \frac{1}{9!}\)

                         đpcm

Tham khảo nhé~

mik không biết khó quá

A = 9/10! + 9/11! + 9/12! +......+ 9/1000! < 9/10! + 10/11! + 11/12! +...+999/1000! = B
9/10! = 1/9! - 1/10!
10/11! = 1/10! - 1/11!
...
999/1000! = 1/999! - 1/1000!
=> B= 1/9! - 1/1000! < 1/9!
=> A < 1/9! \(\left(ĐPCM\right)\).

a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}