- CMR:
\(S1=\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}<\frac{1}{2}\)
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Những câu hỏi liên quan
D
1 tháng 5 2015
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{25}+\frac{1}{41}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{113}\right)\)
< \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{4}{20}+\frac{5}{20}+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)(đpcm)
LT
23 tháng 3 2017
ê cho hỏi tại sao lại ra < \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)
TH
2 tháng 4 2017
Ta có:
\(\hept{\begin{cases}\frac{1}{5}=\frac{1}{5}\\\frac{1}{16}< \frac{1}{5}\\\frac{1}{113}< \frac{1}{5}\end{cases}}...\)\(\Rightarrow\frac{1}{5}+\frac{1}{16}+\frac{1}{25}+\frac{1}{41}+\frac{1}{60}+\frac{1}{85}+\frac{1}{113}< \frac{1}{5}.7=\frac{7}{5}< \frac{10}{5}=2\)(ĐPCM)
11 tháng 12 2019
a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
b) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}=\left(\frac{11}{24}+\frac{13}{24}\right)+\left(-\frac{5}{41}-\frac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
_Học tốt nha_
11 tháng 12 2019
a, \(\frac{15}{12}\)+ \(\frac{5}{13}\)- \(\frac{3}{12}\)-\(\frac{18}{13}\)
= \(\frac{5}{4}\)+ \(\frac{5}{13}\) - \(\frac{1}{4}\) - \(\frac{18}{13}\)
= \(\left(\frac{5}{4}-\frac{1}{4}\right)\)+ \(\left(\frac{5}{13}-\frac{18}{13}\right)\)
= 1 - 1 = 0
b, \(\frac{11}{24}\)- \(\frac{5}{41}\)+ \(\frac{13}{24}\)+ 0,5 - \(\frac{36}{41}\)
= \(\left(\frac{11}{24}+\frac{13}{24}\right)\)- \(\left(\frac{5}{41}+\frac{36}{41}\right)\)+ 0,5
= 1 - 1 + 0,5 = 0,5
c, \(\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)
=\(\left(-\frac{3}{4}+\frac{2}{3}\right).\frac{11}{5}+\left(-\frac{1}{4}+\frac{1}{3}\right).\frac{5}{11}\)
= \(\frac{11}{5}.\left(-\frac{3}{4}+\frac{2}{3}-\frac{1}{4}+\frac{1}{3}\right)\)
= \(\frac{11}{5}.\left[\left(-\frac{3}{4}-\frac{1}{4}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)\right]\)
= \(\frac{11}{5}.\left[\left(-1\right)+1\right]\)
= 0
d, \(\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)
= \(9.\left(0,75-0,25\right)-2\)
= 9. 0,5 - 2 = 2,5
e, \(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
= \(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
= -1 + 1 - \(\frac{1}{2}\)
= \(-\frac{1}{2}\)
29 tháng 6 2022
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
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