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[phankyanh2000]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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[TEX]A1=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{999.1000}[/TEX]
[TEX]A2=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}[/TEX]
[TEX]A3=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}[/TEX]
[TEX]A4=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}[/TEX]
[TEX]A5=\frac{3^3}{10}+\frac{3^3}{40}+\frac{3^3}{88}+...+\frac{3^3}{340}[/TEX]
[TEX]A6=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{31.36}+\frac{36.41}{b}[/TEX]
[TEX]A7=\frac{8}{1.3}+\frac{8}{3.5}+\frac{8}{5.7}+...+\frac{8}{97.99}[/TEX]
[TEX]A8=\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+...+\frac{6}{99}[/TEX]
[TEX]A9=\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}[/TEX]
[TEX]A10=\frac{3}{54}+\frac{5}{126}+\frac{7}{294}+\frac{8}{609}+\frac{9}{1102}[/TEX]

 

[hiensau99]

[TEX]A1=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{999.1000}[/TEX]
[TEX]A_2=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}[/TEX]
[TEX]A_3=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}[/TEX]
[TEX]A_4=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}[/TEX]
[TEX]A_5=\frac{3^3}{10}+\frac{3^3}{40}+\frac{3^3}{88}+...+\frac{3^3}{340}[/TEX]
[TEX]A_6=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{31.36}+\frac{36.41}{b}[/TEX]
[TEX]A_7=\frac{8}{1.3}+\frac{8}{3.5}+\frac{8}{5.7}+...+\frac{8}{97.99}[/TEX]
[TEX]A_8=\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+...+\frac{6}{99}[/TEX]
[TEX]A_9=\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}[/TEX]
[TEX]A_{10}=\frac{3}{54}+\frac{5}{126}+\frac{7}{294}+\frac{8}{609}+\frac{9}{1102}[/TEX]

Làm xong đống này chắc là )
a, $A_1=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{999.1000}$

$=1- \dfrac{1}{2}+ \dfrac{1}{2}- \dfrac{1}{3}+...+\dfrac{1}{999}- \dfrac{1}{1000}$

$=1- \dfrac{1}{1000}= \dfrac{999}{1000}$

b, $A_2=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+ \dfrac{1}{2012.2013}$

$= \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+ \dfrac{1}{2012.2013}$

$= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}- \dfrac{1}{2013}$

$= 1-\dfrac{1}{2013}= \dfrac{2012}{2013}$

c, $A_3=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}$

$=1- \dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}- \dfrac{1}{101}$

$=1- \dfrac{1}{101} = \dfrac{100}{101}$

d, $A_4=\dfrac{3}{10}+\dfrac{3}{40}+\dfrac{3}{88}+...+\dfrac{3}{340}$

$=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{17.20}$

$=\dfrac{1}{2}- \dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}- \dfrac{1}{20}$

$=\dfrac{1}{2}- \dfrac{1}{20}= \dfrac{9}{20}$

e, $A_5=\dfrac{3^3}{10}+\dfrac{3^3}{40}+\dfrac{3^3}{88}+...+\dfrac{3^3}{340}$

$=3^2.(\dfrac{3}{10}+\dfrac{3}{40}+\dfrac{3}{88}+...+\dfrac{3}{340})$

$=3^2. \dfrac{9}{20} = \dfrac{81}{20} $

g, $A_6=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{31.36}+\dfrac{5^2}{36.41}$

$=5.(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{31.36}+\dfrac{5^2}{36.41})$

$=5.(1-\dfrac{1}{6}+\dfrac{1}{6}- \dfrac{1}{11}+...+\dfrac{1}{36}- \dfrac{1}{41})$

$=5.(1- \dfrac{1}{41}) = 5. \dfrac{40}{41} = \dfrac{200}{41}$

h, $A_7=\dfrac{8}{1.3}+\dfrac{8}{3.5}+\dfrac{8}{5.7}+...+\dfrac{8}{97.99}$

$=4.(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99})$

$=4.(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99})$

$=4.(1-\dfrac{1}{99})= 4.\dfrac{98}{99}=\dfrac{392}{99} $

g, $A_8=\dfrac{6}{15}+\dfrac{6}{35}+\dfrac{6}{63}+...+\dfrac{6}{99}$

$A_8=3.(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{9.11})$

$A_8=3.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7} +...+\dfrac{1}{9}- \dfrac{1}{11})$

$A_8=3.(\dfrac{1}{3}- \dfrac{1}{11})= 3.\dfrac{8}{33}=\dfrac{24}{33} $

j, $A_9=\dfrac{1}{2}+\dfrac{2}{2.4}+ \dfrac{3}{4.7}+ \dfrac{4}{7.11}+ \dfrac{5}{11.16}$

$= \dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+ \dfrac{1}{7}- \dfrac{1}{11}+ \dfrac{1}{11}- \dfrac{1}{16}$

$= \dfrac{1}{2}+\dfrac{1}{2}- \dfrac{1}{16}= \dfrac{15}{16}$

k, $A_{10}=\dfrac{3}{54}+\dfrac{5}{126}+\dfrac{7}{294}+ \dfrac{8}{609}+\dfrac{9}{1102}$

$=\dfrac{3}{6.9}+\dfrac{5}{9.14}+\dfrac{7}{14.21}+\dfrac{8}{21.29}+\dfrac{9}{29.38}$

$=\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{29}+ \dfrac{1}{29}-\dfrac{1}{38}$

$= \dfrac{1}{6}- \dfrac{1}{38}= \dfrac{8}{57} $