1/10+1/15+1/21+....+1/105
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C = 3 - 1 + 4 - 1 + 5 - 1 + .... + 102 - 1 + 103 - 1
= 2 + 3 + 4 + ... + 101 + 102
Số số hạng là : (102 - 2) : 1 + 1 = 100 (số hạng)
Tổng : (102 + 2) . 100 : 2 = 5200
Vậy C = 5200
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=\(\dfrac{7}{9}\)
\(\dfrac{45.21.25}{7.15.36}=\dfrac{3.3.3.5.7.5.5}{7.5.3.3.2.3.2}=\dfrac{25}{4}\)
45\7 x 21\15 x 25/36
=45x21x25/7x15x36
=15x3x7x3x25/7x15x3x12
=3x25/12
=75/12
=25/4
a) \(\dfrac{-15}{4}:\dfrac{21}{-10}=\dfrac{-15}{4}.\dfrac{-10}{21}=\dfrac{25}{14}\)
b) \(\dfrac{-7}{14}:\left(-0,14\right)=\dfrac{-7}{14}.\dfrac{-50}{7}=\dfrac{25}{7}\)
c) \(\left(\dfrac{-11}{15}\right):1\dfrac{1}{10}=\dfrac{-11}{15}.\dfrac{10}{11}=\dfrac{-2}{3}\)
d) \(2\dfrac{1}{7}:1\dfrac{1}{14}=\dfrac{15}{7}.\dfrac{14}{15}=2\)
\(a.-\dfrac{15}{4}:\left(\dfrac{21}{-10}\right)\)
\(=-\dfrac{15}{4}\cdot\left(-\dfrac{10}{21}\right)\)
\(=\dfrac{25}{14}\)
\(b.-\dfrac{7}{14}:\left(-0,14\right)\)
\(=-\dfrac{1}{2}:\left(-\dfrac{7}{50}\right)\)
\(=\dfrac{25}{7}\)
\(c.\left(-\dfrac{11}{15}\right):\left(1\dfrac{1}{10}\right)\)
\(=\left(-\dfrac{11}{15}\right):\dfrac{11}{10}\)
\(=-\dfrac{2}{3}\)
\(d.\left(2\dfrac{1}{7}\right):\left(1\dfrac{1}{14}\right)\)
\(=\dfrac{15}{7}:\dfrac{15}{14}\)
\(=2\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{105}+\frac{1}{210}\)
=> \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{210}+\frac{1}{240}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{14.15}+\frac{1}{15.16}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{!}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)
=> \(A=\frac{7}{8}\)
Ta có: \(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=2.\frac{3}{16}=\frac{3}{8}\)
\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}\)
\(=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{210}\right)\)
\(=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+...+\dfrac{1}{14\times15}\right)\)
\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{14}-\dfrac{1}{15}\right)\)
\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{15}\right)\)
\(=2\times\left(\dfrac{15}{60}-\dfrac{4}{60}\right)\)
\(=2\times\dfrac{11}{60}\)
\(=\dfrac{22}{60}=\dfrac{11}{30}\)
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