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OY
20 tháng 7 2021
a) Đặt A= \(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{36}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{2}-\dfrac{1}{9}\)
\(\dfrac{1}{2}\)A=\(\dfrac{7}{18}\)
A=\(\dfrac{7}{9}\)
17 tháng 4 2019
Chứng minh :
\(S=\frac{1}{5}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{25}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{105}< \frac{1}{2}\)
Nhóm các số hạng:
\(S=\frac{1}{5}+\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{25}\right)+\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{105}\right)< \frac{1}{5}+\frac{5}{21}+\frac{5}{101}< \frac{1}{5}+\frac{5}{20}+\frac{5}{100}=\frac{1}{2}.\)
11 tháng 6 2019
\(a,\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+\frac{17}{-35}\)
=\(-\frac{5}{13}+\left(\frac{3}{17}+\frac{14}{17}\right)+\left(\frac{-18}{35}+\frac{-17}{35}\right)\)
= \(-\frac{5}{13}+1+\left(-1\right)\)
=\(-\frac{5}{13}\)
\(b,\frac{-3}{8}.\frac{1}{6}+\frac{3}{-8}.\frac{5}{6}+\frac{-10}{6}\)
=\(\frac{-3}{8}.\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{-10}{6}\)
=\(\frac{-3}{8}.1+\frac{-10}{6}\)
=\(-\frac{49}{24}\)
\(c,\frac{-4}{11}.\frac{5}{15}.\frac{11}{-4}\)
=\(\left(\frac{-4}{11}.\frac{11}{-4}\right).\frac{1}{3}\)
=\(1.\frac{1}{3}=\frac{1}{3}\)
\(d,\frac{13}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}\)
=\(\frac{13}{8}+\frac{1}{2}+\frac{-1}{8}\)
=\(\left(\frac{13}{8}+\frac{-1}{8}\right)+\frac{1}{2}\)
=\(\frac{3}{2}+\frac{1}{2}=2\)
\(e,\frac{-1}{2^2}-\left(-2\right)^2-5\)
=\(\frac{-1}{4}-4-5\)
=\(-\frac{37}{4}\)
\(f,\frac{121}{3}-\frac{5}{7}:\left(24-\frac{23}{57}\right)\)
=\(\frac{121}{3}-\frac{5}{7}:\frac{1345}{57}\)
=\(\frac{121}{3}-\frac{57}{1883}\)
\(\approx40,4\)
PT
1
9 tháng 3 2016
Ta có:\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\frac{10}{39}\)
\(=\frac{5}{39}\)
