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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=\(\dfrac{7}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
(1/3+1/6+1/36) +(1/10+1/15+1/45)+(1/21+1/28)
=(\(\frac{12+6+1}{36}\)+\(\frac{9+6+2}{90}\)+\(\frac{4+3}{84}\)
19/36+17/90+1/12
=(19/36+1/12)+17/90
=7/12+17/90
=105/180+34/180
=139/180
1/3 +1/6+1/10+1/15+1/21+1/28+1/36+1/45
=1/1x3+1/3x2+1/2x5+1/3x5+1/3x7+1/7x4+1/4x9+1/9x5
=1/1-1/3+1/3-1/2....+1/9-1/5
=1/1
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\(P=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\)
\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=2.\dfrac{2}{5}=\dfrac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}\)
\(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\)
\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{9}\)
\(\frac{1}{2}A=\frac{7}{18}\)
\(A=\frac{7}{18}x2\)
\(A=\frac{7}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=2(1/2+1/6+...+1/90)
=2(1-1/2+1/2-1/3+...+1/9-1/10)
=2*9/10=9/5<2
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(2\dfrac{1}{7}:1\dfrac{1}{14}=\dfrac{15}{7}.\dfrac{14}{15}=2\)
\(a.-\dfrac{15}{4}:\left(\dfrac{21}{-10}\right)\)
\(=-\dfrac{15}{4}\cdot\left(-\dfrac{10}{21}\right)\)
\(=\dfrac{25}{14}\)
\(b.-\dfrac{7}{14}:\left(-0,14\right)\)
\(=-\dfrac{1}{2}:\left(-\dfrac{7}{50}\right)\)
\(=\dfrac{25}{7}\)
\(c.\left(-\dfrac{11}{15}\right):\left(1\dfrac{1}{10}\right)\)
\(=\left(-\dfrac{11}{15}\right):\dfrac{11}{10}\)
\(=-\dfrac{2}{3}\)
\(d.\left(2\dfrac{1}{7}\right):\left(1\dfrac{1}{14}\right)\)
\(=\dfrac{15}{7}:\dfrac{15}{14}\)
\(=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
=0,(428571)
HOK TỐT
=0,(428571)
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