Cho Q = 1/30 + 1/42 + 1/56 + 1/72 + 1/90. Khi đó 10.Q = ..................................
cách làm
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6 tháng 3 2018
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
9 tháng 9 2017
Ta có : \(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{9}{10}-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{9.10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\frac{9}{10}=0\)
27 tháng 2 2016
\(Q=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(Q=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\)
\(Q=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\)
Vậy 10Q=1
14 tháng 2 2017
\(Q=\frac{1}{30}+\frac{1}{42}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\)
\(\Rightarrow10Q=\frac{1}{10}.10=1\)
14 tháng 2 2017
\(Q\) = \(\frac{1}{3}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(Q\) = \(\frac{5}{14}+\frac{2}{63}+\frac{1}{90}\)
\(Q\) = \(\frac{7}{18}\)+ \(\frac{1}{90}\)
\(Q=\frac{2}{5}\)
\(10Q=\frac{2}{5}x10\)
\(10Q=4\)
25 tháng 9 2016
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{2}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}\)
\(A=\frac{9}{10}\)(đáp án của p sai nha)
25 tháng 9 2016
= 1 / 1*2 + 1 / 2*3 + 1/ 3*4 + 1 / 4 * 5 ... + 1/ 9*10 = 1-1/2 + 1/2 -1/3 +... + 1/9 - 1/10 = 1 - 1/10 = 9 /10
đáp án của bạn bị sai rùi
S
12 tháng 1 2020
\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}\right)=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{9.10}\right)\)
\(=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...-\frac{1}{10}\right)=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{10}\right)=\frac{1}{5}-\frac{1}{4}=\frac{-1}{20}\)
L
12 tháng 1 2020
\(A=\frac{1}{10}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\)
\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\right)\)
\(A=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(A=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{1}{10}-\left[\left(\frac{1}{4}-\frac{1}{10}\right)-\left(\frac{1}{5}-\frac{1}{5}\right)-...-\left(\frac{1}{9}-\frac{1}{9}\right)\right]\)
\(A=\frac{1}{10}-\frac{1}{4}+\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{4}\)
\(A=-\frac{1}{20}\)
MN
3 tháng 2 2016
\(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(A=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)
\(A=-\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(A=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)=-\left(\frac{1}{4}-\frac{1}{10}\right)=\frac{-3}{20}\)
26 tháng 5 2016
9/1-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2=0/4
26 tháng 5 2016
Giải :
ta có
\(\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
=\(\frac{9}{10}-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}\right)\)
=\(\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{9}{10}-\left[1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\right]\)
=\(\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
=\(\frac{9}{10}-1+\frac{1}{10}=0\) (Mong online math ks cho mình nhé)