Lời giải:
Ta có:
$\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}< \frac{1}{11}+\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}=\frac{10}{11}<1$
Ta có điều phải chứng minh
Chứng tỏ rằng: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}\)
Chứng minh: \(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dots+\dfrac{1}{128}>2\)
Chứng minh rằng: \(\dfrac{1}{201}+\dfrac{1}{202}+\dfrac{1}{203}+\dots+\dfrac{1}{400}< \dfrac{5}{6}\)
Cho S= \(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)
Chứng minh rằng: 1<S<2
\(\dfrac{1}{5}\)+\(\dfrac{1}{13}\)+\(\dfrac{1}{25}\)+...+\(\dfrac{1}{10^2}\)+\(\dfrac{1}{11^2}\)< \(\dfrac{9}{20}\)
Chứng tỏ rằng biểu thức trên bé hơn 9/20
Chứng minh: \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dots+\dfrac{1}{128}>3\)
Chứng minh: \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dots+\dfrac{1}{15}< 3\)
Cho \(C=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{70}\)
Chứng minh rằng : \(\dfrac{4}{3}< C< 2,5\)
Giúp mk vs ..............
1, A= \(\dfrac{-3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
2, B= \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\)
3, C= \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
4, D= \(6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
