CMR: \(\dfrac{1}{3}.\dfrac{4}{6}.\dfrac{7}{9}.\dfrac{10}{12}...\dfrac{208}{210}< \dfrac{1}{25}\)
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bài1
a) \(\dfrac{7}{6}-\dfrac{13}{12}+\dfrac{3}{4}\)
=\(\dfrac{14}{12}-\dfrac{13}{12}+\dfrac{9}{12}\)
=\(\dfrac{1}{12}+\dfrac{9}{12}\)
=\(\dfrac{10}{12}=\dfrac{5}{6}\)
bài 1
b)\(1\dfrac{1}{2}.(\dfrac{-4}{5})\) + \(\dfrac{3}{10}\)
= \(\dfrac{3}{2}.\left(-\dfrac{4}{5}\right)+\dfrac{3}{10}\)
= \(-\dfrac{6}{5}+\dfrac{3}{10}\)
=\(-\dfrac{12}{10}+\dfrac{3}{10}\)
=\(-\dfrac{9}{10}\)
a: 4/9+3/7=28/63+27/63=55/63
3/4+7/24=18/24+7/24=25/24
1/3+2/9+4/27=9/27+6/27+4/27=19/27
b: 5/6-3/8=20/24-9/24=11/24
7/15-11/30=14/30-11/30=3/30=1/10
2/3+1/6-7/12
=8/12+2/12-7/12
=3/12=1/4
c: 18/25*15/6=15/25*18/6=3*3/5=9/5
30/49:6/7=30/49*7/6=210/294=5/7
1/2*3/4:6/5=3/8*5/6=15/48=5/16
d: 8*3/5:12/5=24/5*5/12=2
4:9/5:10/3=4*5/9*3/10=2/3
\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(\dfrac{-44}{21}\right)}\)
\(=\dfrac{53,25+\dfrac{187}{4}}{\dfrac{-25}{11}}\)
\(=\dfrac{100}{\dfrac{-25}{11}}\)
\(=-44\)
Ta có: \(\left(13\frac14-2\frac{5}{27}-10\frac56\right)\cdot230\frac{1}{25}+46\frac34\)
\(=\left(13+\frac14-2-\frac{5}{27}-10-\frac56\right)\cdot230,04+46,75\)
\(=\left(1+\frac14-\frac{5}{27}-\frac56\right)\cdot230,04+46.75\)
\(=\left(\frac{108}{108}+\frac{27}{108}-\frac{20}{108}-\frac{90}{108}\right)\cdot230,04+46,75\)
\(=\frac{25}{108}\cdot230,04+46,75=53,25+46,75=100\)
Ta có: \(\left(1\frac37+\frac{10}{3}\right):\left(12\frac13-14\frac27\right)\)
\(=\left(\frac{10}{7}+\frac{10}{3}\right):\left(12+\frac13-14-\frac27\right)\)
\(=\left(\frac{30}{21}+\frac{70}{21}\right):\left(-2+\frac13-\frac27\right)\)
\(=\frac{100}{21}:\left(-2+\frac{7}{21}-\frac{6}{21}\right)=\frac{100}{21}:\left(-2+\frac{1}{21}\right)=\frac{100}{21}:\frac{-41}{21}=\frac{-100}{41}\)
Ta có: \(\frac{\left(13\frac14-2\frac{5}{27}-10\frac56\right)\cdot230\frac{1}{25}+46\frac34}{\left(1\frac37+\frac{10}{3}\right):\left(12\frac13-14\frac27\right)}\)
\(=100:\frac{-100}{41}=-41\)
a) \(\dfrac{3}{5}+\dfrac{1}{6}=\dfrac{18}{30}+\dfrac{5}{30}=\dfrac{23}{30}\)
b) \(4+\dfrac{2}{5}=\dfrac{20}{5}+\dfrac{2}{5}=\dfrac{22}{5}\)
c) \(\dfrac{25}{12}-\dfrac{7}{6}=\dfrac{25}{12}-\dfrac{14}{12}=\dfrac{11}{12}\)
d) \(\dfrac{9}{7}-\dfrac{7}{6}=\dfrac{54}{42}-\dfrac{49}{42}=\dfrac{5}{42}\)
d) \(\dfrac{3}{7}\times\dfrac{2}{5}=\dfrac{3\times2}{7\times5}=\dfrac{6}{35}\)
Có:
\(\dfrac{n}{n+2}< \dfrac{n-1}{n}\)(Vì
\(n^2< n^2+n-2\forall n>2\))
Nên ta có
\(F=\dfrac{1}{3}.\dfrac{4}{6}....\dfrac{208}{201}\)
\(\Rightarrow F< \dfrac{1}{3}.\dfrac{3}{4}.\dfrac{6}{7}...\dfrac{207}{208}\)
\(\Rightarrow F^2< \dfrac{1.4.7...208}{3.6.9.12...210}.\dfrac{1.3.6.9...207}{3.4.7.10.208}\)
\(\Rightarrow F^2=\dfrac{1}{210}.\dfrac{1}{3}\)
\(\Rightarrow F^2=\dfrac{1}{630}< \left(\dfrac{1}{25}\right)^2\)
Vậy F\(< \dfrac{1}{25}\)