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23 tháng 5 2022
\(=1+1+1+1+1+1+1+1+1+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=9+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
=9+9/10=99/10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 6
a; [6.(- \(\dfrac{1}{3}\))3 - 3.(- \(\dfrac{1}{3}\) + 1)] - ( - \(\dfrac{1}{3}\) - 1)
= [6. \(\dfrac{-1}{3^3}\) - 3.\(\dfrac{2}{3}\)] - ( - \(\dfrac{1}{3}\) - \(\dfrac{3}{3}\))
= [\(\dfrac{-2}{9}\) - 2] + \(\dfrac{4}{3}\)
= [\(\dfrac{-2}{9}\) - \(\dfrac{18}{9}\)] + \(\dfrac{12}{9}\)
= - \(\dfrac{20}{9}\) + \(\dfrac{12}{9}\)
= \(\dfrac{-8}{9}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 6
b; (63 + 3.62 + 33): 13
= (216 + 3.36 + 27) : 13
= (216 + 108 + 27): 13
= (324 + 27): 13
= 351 : 13
= 27
PT
4
HH
19 tháng 6 2018
Giải:
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{1}{1.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}-\dfrac{1}{10.11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}.\dfrac{10}{11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-5}{11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-5}{11}+\dfrac{5}{13}=x\)
\(\Leftrightarrow x=\dfrac{-10}{143}\)
Vậy ...
HH
6
TT
11 tháng 11 2017
\(\dfrac{3}{2}+\dfrac{13}{6}+\dfrac{37}{12}+\dfrac{81}{20}+\dfrac{151}{30}+\dfrac{253}{42}+\dfrac{393}{56}+\dfrac{577}{72}+\dfrac{811}{90}\)
\(=\dfrac{459}{10}\)
LT
1
18 tháng 7 2017
\(\frac{1}{3}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}-\frac{1}{110}=x-\frac{5}{13}\)
\(\frac{1}{3}-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)=x-\frac{5}{13}\)
\(\frac{1}{3}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)=x-\frac{5}{13}\)
\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)=x-\frac{5}{13}\)
\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{11}\right)=x-\frac{5}{13}\)
\(\frac{1}{3}-\frac{1}{3}+\frac{1}{11}=x-\frac{5}{13}\)
\(\frac{1}{11}=x-\frac{5}{13}\)
\(x=\frac{1}{11}+\frac{5}{13}\)
\(x=\frac{68}{143}\)
\(-\frac{3}{20}+\frac{-3}{30}+\frac{-3}{42}+\frac{-3}{56}+\frac{-3}{72}+\frac{-3}{90}\)
\(=-3\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=-3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=-3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-3\left(\frac{1}{4}-\frac{1}{10}\right)=-\frac{9}{20}\)
\(\frac{-3}{20}+\frac{-3}{30}+\frac{-3}{42}+\frac{-3}{56}+\frac{-3}{72}+\frac{-3}{90}\)
\(=-3\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=-3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{9\cdot10}\right)\)
\(=-3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-3\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=-3\cdot\frac{3}{20}\)
\(=\frac{-9}{20}\)