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1)
a) Do 5/5 = 1
=> 1/5 < 1
Do 6/6 = 1
=> 7/6 > 1
=> 7/6 > 1/5
b) Như trên ta có : 3/7 < 1
4/2 > 1
=> 4/2 > 3/7
2)
a ) <
b) >
c) =
Ta có :
2018 x 2018 = ( 2017 + 1 ) x ( 2019 - 1 )
= ( 2017 + 1 ) x 2019 - ( 2017 + 1 )
= 2017 x 2019 + 2019 - 2017 - 1
= 2017 x 2019 + 1 > 2017 x 2019
\(\Rightarrow\frac{2018\times2018}{2017\times2019}=\frac{2017\times2019+1}{2017\times2019}=1+\frac{1}{2017\times2019}>1\)
Vậy ta chọn B
~~Học tốt~~
a, \(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{2011\cdot2011}\)
có :
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{2011\cdot2011}< \frac{1}{2010\cdot2011}\)
nên :
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2010\cdot2011}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(\Rightarrow A< 1-\frac{1}{2011}\)
\(\Rightarrow A< \frac{2010}{2011}< 1\)
b, \(A=\frac{2010}{2011}=1-\frac{1}{2011}\)
\(\frac{3}{4}=1-\frac{1}{4}\)
\(\frac{1}{4}>\frac{1}{2011}\)
nên :
\(A>\frac{3}{4}\)
\(Giải\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)\(+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2014}\)
\(A=0+0+0+...+0+0\)
\(\Rightarrow A=0\)
\(a.\)\(A< 1\)
b. \(A< \frac{3}{4}\)
8/9 = 8x3/9x3=24/27
12/13 = 12x2/13x2=24/26
Vậy 24/26 > 24/27
12/36 = 12x5/35x5 = 60/175
15/60=15x4/60x4 = 60/240
60/175 > 60/240 nên 12/36 > 15/60
\(a.\frac{8}{9}và\frac{12}{13}\)
\(\frac{8}{9}=\frac{8\times3}{9\times3}=\frac{24}{27}vả\frac{12}{13}=\frac{12\times2}{13\times2}=\frac{24}{26}\)
\(Vì\frac{24}{27}< \frac{24}{26}nên\frac{8}{9}< \frac{12}{13}\)
\(b.\frac{12}{36}và\frac{15}{60}\)
\(\frac{12}{36}=\frac{12\div4}{36\div4}=\frac{3}{9}và\frac{15}{60}=\frac{15\div5}{60\div5}=\frac{3}{12}\)
\(Vì\frac{3}{9}>\frac{3}{12}nên\frac{12}{36}>\frac{15}{60}\)
Quy đồng tử số : 6/7 = 6 x20/7 x 20 = 120/140 . Vì 140 lớn hơn 137 nen 120/140 < 120/137 hay 6/7 < 120/137 .Vay 6/7 < 120/37.