CMR:
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{32}>2\)
KO hieu vao phan "Doc them"
Tinh:
\(\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
Giup minh trinh bay cach cach giai voi nhe
Ko hieu vao phan"Doc them"
\(\frac{\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)=\(\frac{1}{\frac{7}{2}}\)=\(\frac{2}{7}\)
Tinh
D= \(\frac{\frac{1}{3}+\frac{1}{17}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{17}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
Trinh bay cach lam nhe
Ko hieu vao DOC THEM
Tinh:
\(B=\frac{\frac{1}{7}+\frac{1}{23}-\frac{1}{1009}}{\frac{1}{7}+\frac{1}{23}-\frac{1}{1009}+\frac{1}{7}.\frac{1}{23}.\frac{1}{1009}}\)
Trinh bay cach lam nua nhe
Co gi ko hieu thi vao phan doc them
Tinh :
A=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.1016}\right)\)
Giup minh trinh bay ca cach lam nua nha!
Ai chua doc het de va ko hieu thi vao phan "Doc them" se ro
Thanks
Tinh :
A=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)......\left(1+\frac{1}{2014.2016}\right)\)
Giup minh trinh bay ca cach lam nua nha!
Ai chua doc het de va ko hieu thi vao phan "Doc them" se ro
Thanks
Chung minh:
\(\frac{n\left(n+1\right)}{2}\) va 2n+1 nguyen to cung nha voi moi n thuoc N
helps me
Neu ai doc ko het thi vao phan doc them thi se hieu ro hon nhe
So sanh A va B biet: A=\(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}\) va B=\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{17}\)
Trinh bay cach lam cho minh voi nhe
Ai ko hieu thi vao phan "Doc them"
Giup minh voi nhe!Please
tim x biet:
\(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}\)
trinh bay cach giai ho minh nhe
co gi ko hieu thi vao cho"doc them"
a) Cho \(s=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
CMR 1<s<2, từ đó suy ra s ko phải stn
b) Cho \(s=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+....+\frac{1}{60}\)
CMR 3/5< s < 4/5
a)ta có:
\(\frac{3}{10}\)>\(\frac{3}{15}\)
\(\frac{3}{11}\)>\(\frac{3}{15}\)
...
\(\frac{3}{14}\)>\(\frac{3}{15}\)
Cộng từng vế của bất đẳng thức trên ta được:
\(\frac{3}{10}\)+\(\frac{3}{11}\)+\(\frac{3}{12}\)+\(\frac{3}{13}\)+\(\frac{3}{14}\)<\(\frac{3}{15}\)+\(\frac{3}{15}\)+\(\frac{3}{15}\)+\(\frac{3}{15}\)+\(\frac{3}{15}\)
Hay S>\(\frac{15}{15}\)=>S>1 (1)
ta có :
\(\frac{3}{11}\)<\(\frac{3}{10}\)
\(\frac{3}{12}\)<\(\frac{3}{10}\)
...
\(\frac{3}{14}\)<\(\frac{3}{10}\)
Cộng từng vế của bất đẳng thức trên ta được:
\(\frac{3}{10}\)+\(\frac{3}{11}\)+\(\frac{3}{12}\)+\(\frac{3}{13}\)+\(\frac{3}{14}\)<\(\frac{3}{10}\)+\(\frac{3}{10}\)+\(\frac{3}{10}\)+\(\frac{3}{10}\)+\(\frac{3}{10}\)
Hay S<\(\frac{15}{10}\)<\(\frac{20}{10}\)=2
Vậy S<2 (2)
Theo câu 1 ta có : S>1
Theo câu 2 ta có :S<2
Vậy 1<S<2
=>S ko phải số tự nhiên