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6 tháng 7 2019
a)\(\frac{1}{2}-2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{50}\right)\)
=\(\frac{1}{50}\)
BK
6 tháng 7 2019
\(1)a)\frac{1}{2}-2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{24.25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{24}-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\frac{24}{25}=\frac{-23}{50}\)
\(\)
XO
26 tháng 9 2019
A = \(\frac{1}{3}+\frac{13}{35}+\frac{33}{35}+\frac{61}{63}+\frac{97}{99}+\frac{141}{143}\)
\(=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)\)
\(=\left(1+1+1+1+1+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(=6-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=6-\left(1-\frac{1}{13}\right)\)
\(=6-1+\frac{1}{13}\)
\(=5+\frac{1}{13}\)
\(=\frac{66}{13}\)
\(\text{Vậy }A=\frac{66}{13}\)
31 tháng 7 2015
Ta có: \(A=\frac{3-2}{3}+\frac{15-2}{15}+\frac{35-2}{35}+\frac{63-2}{63}+\frac{99-2}{99}+\frac{143-2}{143}+\frac{195-2}{195}\)
\(A=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)+\left(1-\frac{2}{195}\right)\)
\(A=7-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}+\frac{2}{195}\right)\)
\(A=7-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(A=7-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(A=7-\left(1-\frac{1}{15}\right)=7-1+\frac{1}{15}=6\frac{1}{15}\)không là số nguyên
PL
6 tháng 3 2020
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
4 tháng 11 2018
\(M=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right).....\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)=\frac{-3}{4}.\frac{-8}{9}.....\frac{-99}{100}.\frac{-120}{121}\)
\(M=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.....\frac{-9.11}{10.10}.\frac{-10.12}{11.11}=\frac{-1}{2}.\frac{-12}{11}=\frac{12}{22}=\frac{6}{11}\)
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{99}\)
\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{9.11}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(S=\frac{5}{11}\)
\(Q=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2013.2015}\)
\(Q=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(Q=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(Q=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)
\(Q=\frac{1007}{2015}\)
~ Đấng Ed :) ~
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
17 tháng 7 2019
Bài 1c)
\(\frac{1}{3}+x=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\)
\(\frac{1}{3}+x=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)
\(\frac{1}{3}+x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{3}+x=\frac{1}{3}-\frac{1}{11}=\frac{11}{33}-\frac{3}{33}=\frac{8}{33}\)
\(x=\frac{8}{33}-\frac{1}{3}=\frac{8}{33}-\frac{11}{33}=\frac{-3}{33}=\frac{-1}{11}\)
HT
25 tháng 7 2017
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)
\(\Rightarrow3x-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)=0\)
\(\Rightarrow3x-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)=0\)
\(\Rightarrow3x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)
\(\Rightarrow3x-\left(1-\frac{1}{99}\right)=0\)
\(\Rightarrow3x-\frac{98}{99}=0\)
\(\Rightarrow3x=0+\frac{98}{99}\)
\(\Rightarrow3x=\frac{98}{99}\)
\(\Rightarrow x=\frac{98}{99}:3\)
\(\Rightarrow x=\frac{98}{297}\)
\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)
\(2\left(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}\right)=2.0\)
\(6x-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}-\frac{2}{99}=0\)
\(6x-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=0\)
\(6x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)
\(6x-\left(1-\frac{1}{11}\right)=0\)
\(6x-\frac{10}{11}=0\)
\(6x=\frac{10}{11}\)
\(x=\frac{5}{33}\)