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a) \(\dfrac{23}{24}< 1\)
\(\dfrac{24}{23}>1\)
\(\Rightarrow\dfrac{23}{24}< \dfrac{24}{23}\)
b) \(\dfrac{4}{21}< \dfrac{4}{20}=\dfrac{1}{5}=\dfrac{6}{30}< \dfrac{6}{29}\)
c) \(\dfrac{6}{7}=1-\dfrac{1}{7}< \dfrac{8}{9}=1-\dfrac{1}{9}\)
d) \(\dfrac{1212}{1313}=\dfrac{12\times101}{13\times101}=\dfrac{12}{13}\)
M = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024
4.M = 1 + 1/4 + 1/16 + 1/64 + 1/256
4M - M = (1 + 1/4 + 1/16 + 1/64 + 1/256 ) - ( 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 )
3M = 1 - 1/1024
3M = 1023/1024
M = 341/1024
M=\(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{1024}\)
=\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\)
=>4M=1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)
=>4M-M=3M=(1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\))-(\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\))=1-\(\dfrac{1}{4^5}\)=\(\dfrac{1023}{1024}\)
=>M=\(\dfrac{1023}{1024}\):3=\(\dfrac{341}{1024}\)
a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh
a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x-\dfrac{23}{14}=1\)
\(x=1+\dfrac{23}{14}\)
\(x=\dfrac{37}{14}\)
b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{28}{15}\)
a)\(\dfrac{19}{10}>\dfrac{10}{11}\)
b)\(\dfrac{11}{10}=\dfrac{12}{11}\)
c)\(\dfrac{9}{10}< \dfrac{10}{11}\)
a) \(\dfrac{27}{35}>\dfrac{19}{35}>\dfrac{19}{41}\)
\(\Rightarrow\dfrac{27}{35}>\dfrac{19}{41}\)
b) \(\dfrac{120}{121}< \dfrac{120+1}{121+1}=\dfrac{121}{122}\)
\(\Rightarrow\dfrac{120}{121}< \dfrac{121}{122}\)
a
2/5> 2/7
5/9<5/6
11/2>11/3
cách so sánh :
sét mẫu số của phân số này bé hơn mẫu số của phân số kia thì phân số này lớn hơn
mẫu số của phân số này lớn hơn mẫu số của phân số kia thì phân số này bé hơn
`3/7-2/5`
`=1/35>0`
`=>3/7>2/5`
`b,9>8`
`=>1/9<1/8`
`=>5/9<5/8`
`d,8/7>1`
`7/8<1`
`=>8/7>7/8`
1999/1990 và 2000/1991
1999/1990-1=9/1990
2000/1991-1=9/1991
Vì 9/1990>9/1991 nên 1999/1990 < 2000/1991
2780/2770 và 2555/2550
2780/2770 -1=1/277
2555/2550-1=1/510
Vì 1/277>1/510 nên 2780/2770 < 2555/2550
8089/8080 và 9879/9870
8089/8080-1=9/8080
9879/9870-1=3/3290=9/9870
Vì 9/8080> 9/9870 nên 8089/8080< 9879/9870
`a)1<3`
`=>1/5<3/5`
`b)21>9`
`=>8/21<8/9`
`c)3/5<5/5=1`
`d)7/5>5/5=1`
A = \(\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)
Ta có:
\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)
Từ 3 điều trên suy ra : A < B
\(\dfrac{774}{763}\) > \(\dfrac{774}{768}\) = \(\dfrac{129}{128}\) = 1 + \(\dfrac{1}{128}\) > 1 + \(\dfrac{1}{256}\) = \(\dfrac{257}{256}\)
vậy \(\dfrac{774}{763}\) > \(\dfrac{257}{256}\)
b, \(\dfrac{257}{256}\) = 1 + \(\dfrac{1}{256}\) > 1 + \(\dfrac{1}{773}\) = \(\dfrac{774}{773}\)
vậy \(\dfrac{257}{256}\) > \(\dfrac{774}{773}\)