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RT
2
11 tháng 7 2021
a) Ta có: \(\dfrac{1}{2022}-\dfrac{5}{2\cdot4}-\dfrac{5}{4\cdot6}-\dfrac{5}{6\cdot8}-...-\dfrac{5}{2020\cdot2022}\)
\(=\dfrac{1}{2022}-5\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\cdot\dfrac{1010}{2022}\)
\(=\dfrac{1}{2022}-\dfrac{2025}{2022}=\dfrac{-1262}{1011}\)
11 tháng 7 2021
b) Ta có: \(\dfrac{2^2}{1\cdot3}+\dfrac{2^2}{3\cdot5}+...+\dfrac{2^2}{197\cdot199}\)
\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)
\(=2\left(1-\dfrac{1}{199}\right)\)
\(=2\cdot\dfrac{198}{199}=\dfrac{396}{199}\)
TD
4
N
9 tháng 9 2019
\(=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
9 tháng 9 2019
5/1.3 + 5/3.5 + ... + 5/99.101
= 5/2.(1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)
= 5/2.(1 - 1/101)
=5/2.100/101
= 250/101
SY
1
29 tháng 8 2020
a) 8.(-5).(-4).2 = 8.20.2 = 8.40 = 320
b) \(1\frac{3}{7}+\left(-\frac{1}{3}+2\frac{4}{7}\right)\)
\(=1\frac{3}{7}-\frac{1}{3}+2\frac{4}{7}\)
\(=\left(1\frac{3}{7}+2\frac{4}{7}\right)-\frac{1}{3}=\left(1+2\right)+\left(\frac{3}{7}+\frac{4}{7}\right)-\frac{1}{3}=4-\frac{1}{3}=\frac{11}{3}\)
c) \(\frac{8}{5}\cdot\frac{-2}{3}+\frac{-5\cdot5}{3\cdot5}\)
\(=\frac{8}{5}\cdot\frac{-2}{3}+\frac{-25}{15}=\frac{-16}{15}+\frac{-25}{15}=\frac{-41}{15}\)
d) \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\left(-2\right)^2=\frac{6}{7}+\frac{5}{8}\cdot\frac{1}{5}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{13}{56}\)
NT
4
IM
13 tháng 8 2016
Đặt biểu thức là A
=> \(A=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\right)\)
=> \(A=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
=> \(A=\frac{5}{2}.\frac{100}{101}\)
=> \(A=\frac{250}{101}\)
13 tháng 8 2016
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
LA
2
10 tháng 12 2018
a,
\(\Rightarrow5^x.(1+5)=150\)
\(\Rightarrow5^x.6=150\)
\(\Rightarrow5^x=150:6\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
LL
1
PA
30 tháng 7 2016
\(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
NT
1
2 tháng 2 2022
\(=\dfrac{5^6\cdot3^6\cdot\dfrac{1}{2}+5^8}{5^6\cdot26}=\dfrac{3^6\cdot\dfrac{1}{2}+5^2}{26}=\dfrac{779}{52}\)
\(\left(\frac{-1}{5}\right)^5\cdot\left(\frac{3}{5}\right)^5=\left(\frac{-1}{5}\cdot\frac{3}{5}\right)^5=\left(\frac{-3}{25}\right)^5\)
mk nhé!
\(\left(-\frac{1}{5}\right)^5\cdot\left(3\cdot5\right)^5\)
\(=\left(-\frac{1}{5}\cdot3\cdot5\right)^5\)
\(=-3^5\)
\(=-243\)
ti ck nha