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MT
28 tháng 3 2020
Đặt \(A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+...+\frac{3}{9900}\)
\(=\frac{3}{1\times2}+\frac{3}{2\times3}+\frac{3}{3\times4}+...+\frac{3}{99\times100}\)
\(\Rightarrow A:3=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{100}\times3=\frac{297}{100}\)
Vậy \(A=\frac{297}{100}\).
P
5
10 tháng 4 2019
\(=3.\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)\)
\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
10 tháng 4 2019
\(\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+...+\frac{3}{9900}\\ =3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=3\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{100-99}{99.100}\right)\\ =3\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)=3.\frac{99}{100}=\frac{297}{100}\)
LH
1
26 tháng 8 2016
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(=1-\frac{1}{5}\)
\(=\frac{4}{5}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
7 tháng 8 2015
a) \(\frac{13}{7}-\frac{1}{2}\times\frac{13}{7}+\frac{3}{2}\times\frac{13}{7}\)
\(=\frac{13}{7}\times\left(1-\frac{1}{2}+\frac{3}{2}\right)\)
\(=\frac{13}{7}\times2\)
\(=\frac{26}{7}\)
b) \(\frac{1}{15}\times\left(\frac{3}{7}+\frac{5}{19}\right)+\frac{3}{7}\times\left(\frac{5}{19}-\frac{1}{15}\right)\)
\(=\frac{1}{15}\times\frac{3}{7}+\frac{1}{15}\times\frac{5}{19}+\frac{3}{7}\times\frac{5}{19}-\frac{3}{7}\times\frac{1}{15}\)
\(=\frac{5}{19}\times\left(\frac{1}{15}+\frac{3}{7}\right)\)
\(=\frac{5}{19}\times\frac{52}{105}\)
\(=\frac{52}{399}\)
c) \(\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}\)
\(=5\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{99\times100}\right)\)
\(=5\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5\times\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=5\times\frac{49}{100}\)
\(=\frac{49}{20}\)
Lần sau nên đăng ít thôi
C
0
TC
10 tháng 7 2020
Ta có:
\(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{9901}{9900}\)
\(A=1+\frac{1}{2}+1+\frac{1}{6}+1+\frac{1}{12}+...+1+\frac{1}{9900}\)\(A=1+1+1+...+1(51c/s)+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(A=51+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=51+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=51+1-\frac{1}{100}\)
\(A=52-\frac{1}{100}\)
\(A=\frac{5199}{100}\)
Cái đoạn 1+1+1+...+1 ( 51 c/s) số tớ ko thể giải thích trên máy tính đc nên bn tự suy nghĩ nhé:)))
KT
15 tháng 7 2018
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Mình chỉnh lại đề B nha:
\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
15 tháng 7 2018
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
2 tháng 7 2016
\(F=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+......+\frac{3}{870}\)
\(=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+.....+\frac{3}{29.30}\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{29}-\frac{1}{30}\right)\)
\(=3.\left(1-\frac{1}{30}\right)\)
\(=3.\frac{29}{30}\)
\(=\frac{29}{10}\)
nhanh nha mình đang cần