CMR:
B=1/12+1/13+...+1/22 < 1/2
D=1/5+1/6+....+1/17 < 2
HELP ME !!!!!
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a,1/51 > 1/100
1/52 > 1/100
1/53 > 1/100
...
1/100=1/100
=>H>1/100 + 1/100 + 1/100 +...+1/100
H>50/100=1/2
1/51<1/50
1/52<1/50
....
1/100<1/50
=>H<1/50+1/50+...+1/50
H<50/50=1
Vay1/2<H<1
Lời giải:
a, Ta có: \(A=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+...+\frac{1}{22}>\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=\frac{1}{22}.11=\frac{11}{22}=\frac{1}{2}\)
Vậy: \(A>\frac{1}{2}\)
b, Ta có: \(B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\)
\(=\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Mà: \(\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\text{}\text{}\text{}>\left(\frac{1}{50}+...+\frac{1}{50}+\frac{1}{50}\right)+\left(\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\right)\)
=> \(B\text{}\text{}\text{}>\frac{1}{50}.41+\frac{1}{100}.50=\frac{41+25}{50}=\frac{33}{25}>1\)
Vậy: \(B>1\)
c, Ta có: \(C=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}< \frac{1}{5}+\frac{1}{6}+\left(\frac{1}{7}+...+\frac{1}{7}+\frac{1}{7}\right)=\frac{11}{30}+11.\frac{1}{7}=\frac{407}{210}< \frac{420}{210}=2\)
Vậy: \(C< 2\)
Chúc bạn học tốt!Tick cho mình nhé!
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}.\)
\(B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(B< 1-\frac{1}{8}< 1\)
1)\(\frac{252}{x}=\frac{84}{97}\Rightarrow\)\(\frac{84}{97}=\frac{252}{291}\Rightarrow x=291\)
6) \(\frac{y}{15}=\frac{2}{5}\Rightarrow\frac{2}{5}=\frac{6}{15}\Rightarrow x=6\)
Hình như sai đề :) T sửa lại nhé
\(B=\frac{1}{12}+\frac{1}{13}+...+\frac{1}{22}>\frac{1}{2}\)
B có 11 số hạng
Ta có: \(\frac{1}{12}>\frac{1}{22}\)
\(\frac{1}{13}>\frac{1}{22}\)
............
\(\frac{1}{22}=\frac{1}{22}\)
\(\Rightarrow B>\left(\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}\right)=\frac{11}{22}=\frac{1}{2}\)
\(D=\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}< 2\)
Ta có: \(\frac{1}{5}=\frac{1}{5};\frac{1}{6}< \frac{1}{5};...;\frac{1}{10}< \frac{1}{5}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}< (\frac{1}{5}+\frac{1}{5}+...+\frac{1}{5})=\frac{6}{5}\)(1)
Lại có: \(\frac{1}{11}=\frac{1}{11};\frac{1}{12}< \frac{1}{11};\frac{1}{13}< \frac{1}{11};...;\frac{1}{17}< \frac{1}{11}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{17}< (\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11})=\frac{7}{11}\)(2)
Từ (1), (2) \(\Rightarrow D< \frac{6}{5}+\frac{7}{11}=\frac{101}{55}< \frac{110}{55}=2\)
P/s: Hoq chắc :<