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Ta có:
\(\dfrac{1}{5}>\dfrac{1}{10}\\ \dfrac{1}{6}>\dfrac{1}{10}\\ ...\\ \dfrac{1}{9}>\dfrac{1}{10}\\ \Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}>\dfrac{5}{10}=\dfrac{1}{2}.\)
Tương tự:
\(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}>\dfrac{5}{15}=\dfrac{1}{3}.\\ \dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}>\dfrac{3}{18}=\dfrac{1}{6}.\)
Cộng vế theo vế ta được \(B>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\left(đpcm\right)\)
->1/1001 +1/1002 +...+ 1/2000 < 1/2000 + 1/2000+...+ 1/2000(1000 lần 1/2000 vì 1000 là số số hạng từ 1001 đến 2000, hiểu ý mình chứ) Mà 1/2000 * 1000 = 1000/2000 =1/2<3/4 =>1/1001 + 1/1002 +...+ 1/2000>3/4
Merry Christmas!!!!!!!
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)
Ta có:\(\frac{1}{2}>\frac{1}{8};\frac{1}{3}>\frac{1}{8};...;\frac{1}{6}>\frac{1}{8};\frac{1}{7}+\frac{1}{8}+\frac{1}{9}>\frac{3}{8}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{3}{8}\)
\(=\frac{8}{8}=1\)
Vậy\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>1\)
Gọi phép tính 1^2+2^2+6^2 là A;phép tính 2^2+3^2+7^2 là B
*1^2=1
5^2=25
6^2=36
Vậy A=1^2+5^2+6^2=1+25+36=62
*2^2=4
3^2=9
7^2=49
Vậy B= 2^2+3^2+7^2=4+9+49=62
Suy ra A=B nhé
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{29}+\frac{1}{30}\)
\(A=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\right)\)
\(A>\left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)\)
\(A>10.\frac{1}{20}+10.\frac{1}{30}\)
\(A>\frac{1}{2}+\frac{1}{3}\)
\(A>\frac{5}{6}\)
Vậy \(A>\frac{5}{6}\)
Chúc bạn học tốt ~
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{29}+\frac{1}{30}\)
\(A=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\right)\)
\(A>\left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)\)
\(A>\frac{1}{20}\times10+\frac{1}{30}\times10\)
\(A>\frac{1}{2}+\frac{1}{3}\)
\(A>\frac{5}{6}\)
Vậy \(A>\frac{5}{6}\)
Cái của mình bị sai. Để mình làm lại nhé!
Ta có: \(\dfrac{1}{5}=\dfrac{1}{5}\)
\(\dfrac{1}{6}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{10}\)
...
\(\dfrac{1}{10}< \dfrac{1}{5}\)
Do đó: \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{10}< \dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{5}=\dfrac{6}{5}\)
Ta có: \(\dfrac{1}{11}=\dfrac{1}{11}\)
\(\dfrac{1}{12}< \dfrac{1}{11}\)
...
\(\dfrac{1}{17}< \dfrac{1}{11}\)
Do đó: \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{17}< \dfrac{1}{11}+\dfrac{1}{11}+...+\dfrac{1}{11}=\dfrac{7}{11}\)
Vì \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{10}< \dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{5}=\dfrac{6}{5}\)
và \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{17}< \dfrac{1}{11}+\dfrac{1}{11}+...+\dfrac{1}{11}=\dfrac{7}{11}\)
nên \(P< \dfrac{6}{5}+\dfrac{7}{11}=\dfrac{101}{55}< \dfrac{110}{55}=2\)
hay P<2