\(\left(1-\frac{1}{1007}\right)\left(1-\frac{1}{1008}\right)\left(1-\frac{1}{1009}\right)\left(1-\frac{1}{1010}\right)\left(1-\frac{1}{1011}\right)\left(1-\frac{1}{1012}\right)\)

\(=\frac{1006}{1007}\cdot\frac{1007}{1008}\cdot\frac{1008}{1009}\cdot\frac{1009}{1010}\cdot\frac{1010}{1011}\cdot\frac{1011}{1012}\)

\(=\frac{1006\cdot1007\cdot1008\cdot1009\cdot1010\cdot1011}{1007\cdot1008\cdot1009\cdot1010\cdot1011\cdot1012}=\frac{503}{506}\)

=\(\frac{1006}{1007}.\frac{1007}{1008}.....\frac{1011}{1012}\)

=\(\frac{1006}{1012}\)

=\(\frac{503}{506}\)

nếu sai sót mong mọi người sửa lỗi đúng thì ủng hộ

\(\left(1-\frac{1}{1007}\right)\left(1-\frac{1}{1008}\right)\left(1-\frac{1}{1009}\right)...\left(1-\frac{1}{1012}\right)\)

\(=\frac{1006}{1007}\cdot\frac{1007}{1008}\cdot\frac{1008}{1009}\cdot...\cdot\frac{1011}{1012}\)

\(=\frac{1006\cdot1007\cdot1008\cdot...\cdot1011}{1007\cdot1008\cdot1009\cdot...\cdot1012}=\frac{1006}{1012}=\frac{503}{506}\)

Ta có: \(\left(1-\frac{1}{1007}\right)\times\left(1-\frac{1}{1008}\right)\times...\times\left(1-\frac{1}{1011}\right)\times\left(1-\frac{1}{1012}\right)\)

\(=\frac{1006}{1007}\times\frac{1007}{1008}\times...\times\frac{1010}{1011}\times\frac{1011}{1012}\)

\(=\frac{1006}{1012}=\frac{503}{506}\)

\(\left(1-\frac{1}{1007}\right)\cdot\left(1-\frac{1}{1008}\cdot\right)...\cdot\left(1-\frac{1}{1011}\right)\cdot\left(1-\frac{1}{1012}\right)\)

\(=\frac{1006}{1007}\cdot\frac{1007}{1008}\cdot...\cdot\frac{1010}{1011}\cdot\frac{1011}{1012}\)

\(=\frac{1006.1007\cdot..\cdot2010\cdot2011}{1007\cdot1008\cdot....\cdot1011.1012}\)

\(=\frac{1006}{1012}\)

\(=\frac{503}{506}\)

Cảm ơn m.n nha, ko ngờ lại có kết quả nhanh như vậy(=^-ω-^=)

Ta có : 

\(S=\left(1+\frac{1}{3}+..+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)

\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}=P\)

\(\Rightarrow\left(s-p\right)^{2013}=0^{2013}=0\)

Thanks bạn nhiều nhé!! Tặng bạn 1 tk :>

Mà bạn ơi làm giúp mình bài này nữa đc ko?? 

Ta có : S =\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)

\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\)\(-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)\(-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)

\(\Rightarrow S=P\)

Khi đó : \(\left(S-P\right)^{2018}=0^{2018}=0\)

k chi mik nha!

-.-

A=1+1/2+1/3+1/4+...+1/2^2018-1 Chứng tỏ A<2018

Giải:

A=10^11-1/10^12-1

10A=10.(10^11-1)/10^12-1

10A=10^12-10/10^12-1

10A=10^12-1-9/10^12-1

10A=10^12-1/10^12-1 + -9/10^12-1

10A=1+ -9/10^12-1

B=10^10+1/10^11+1

10B=10.(10^10+1)/10^11+1

10B=10^11+10/10^11+1

10B=10^11+1+9/10^11+1

10B=10^11+1/10^11+1 + 9/10^11+1

10B=1 + 9/10^11+1

Vì -9/10^12-1 < 9/10^11+1 nên 10A < 10B

=>A < B

Chúc bạn học tốt!