(1+(4)/(5))(1+(4)/(12))(1+(4)/(21))..(1+(4)/(n(n+4)))<6
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29 tháng 8 2017
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
LN
18 tháng 9 2018
a,=\(\dfrac{8}{14}-\dfrac{1}{14}+\dfrac{5}{21}+\dfrac{3}{2}\)
=\(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{5}{21}\) =\(2+\dfrac{5}{21}\) =\(\dfrac{42}{21}+\dfrac{5}{21}\) =\(\dfrac{47}{21}\)
b,=\(\dfrac{11}{13}.\dfrac{12}{15}-\dfrac{7}{15}+\dfrac{14}{15}.\dfrac{11}{13}\)
=\(\dfrac{11}{13}.\left(\dfrac{12}{15}+\dfrac{14}{15}\right)-\dfrac{7}{15}\)
=\(\dfrac{11}{13}.\dfrac{26}{15}-\dfrac{17}{15}\) =\(\dfrac{22}{15}-\dfrac{17}{15}\) =\(\dfrac{5}{15}\) =\(\dfrac{1}{3}\)
c,=\(\left(\dfrac{3}{6}-\dfrac{2}{6}\right)^2\) =\(\left(\dfrac{1}{6}\right)^2\) =\(\dfrac{1}{36}\)
d,=câu này dễ mà
24 tháng 3 2019
a) \(\frac{-1}{21}+\frac{-1}{28}-\frac{-1}{21}-\frac{3}{14}\)
\(=\frac{-1}{21}+\frac{-1}{28}+\frac{1}{21}+\frac{-3}{14}\)
\(=\left(\frac{-1}{21}+\frac{1}{21}\right)+\left(\frac{-1}{28}+\frac{-3}{14}\right)\)
\(=0+\left(\frac{-1}{28}+\frac{-6}{28}\right)\)
\(=0+\frac{-1}{4}=\frac{-1}{4}\)
b) \(\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(=\frac{-1}{5}+\frac{1}{4}+\frac{-3}{4}\)
\(=\left(\frac{1}{4}+\frac{-3}{4}\right)+\frac{-1}{5}\)
\(=\frac{-1}{2}+\frac{-1}{5}\)
\(=\frac{-5}{10}+\frac{-2}{10}=\frac{-7}{10}\)
c) \(\frac{-4}{11}\cdot\frac{2}{5}+\frac{6}{11}\cdot\frac{-3}{10}\)
\(=\frac{-8}{55}+\frac{-9}{55}=\frac{-17}{55}\)
14 tháng 9 2017
a) 3/4 + -1/8 = 5/8
b)-5/12 + -7/24 = -9/8
c) 4/21 - -5/28 = 31/84
d) 1 + -7/28 = 3/4
e) -4/3 - 17/6= -25/6
f) 1/3 - ( 1/2 +1/8 )= -7/24
g)1/21 - ( 1/7 - 1/3 ) = 5/21
h)1/2 - 1/4 + 1/13 + 1/8= 47/104
14 tháng 9 2017
a) x - 1/10 = 1/15
x=1/15+1/10
x=1/6 Vay x=1/6b) -4/21 - x = -3/7
x=-4/21+3/7 x=5/21 Vay x=5/21c) x + 1/2 = 3/4 - (-1/2)
x+1/2= 5/4
x= 5/4-1/2
x=3/4
Vay x=3/4
d) 4/7 - x = 1/3 - (-2/3)
x= 4/7-1/3-2/3 x= -3/7 Vay x=-3/7
ND
2
3 tháng 12 2017
Sửa đề : S= -1/2-1/3-1/4-.....-1/20 + 3/2 + 4/3 + 5/4 + ... + 21/20 . Tính S
\(S=\left(\frac{3}{2}-\frac{1}{2}\right)+\left(\frac{4}{3}-\frac{1}{3}\right)+\left(\frac{5}{4}-\frac{1}{4}\right)+...+\left(\frac{21}{20}-\frac{1}{20}\right)\)
\(S=1+1+1+...+1\)( 20 số 1 )
\(S=20\)
F
23 tháng 1 2020
Bài giải
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
M
23 tháng 1 2020
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)