\(\dfrac{1}{2010}và\dfrac{-7}{19}\)
so sánh
K
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1 tháng 4 2021
A=-2015/2015x2016
A=-1/2016
B=-2014/2014x2015
B=-1/2015
vi 2016>2015,-1/2016>-1/2015
vay A>B
1 tháng 4 2021
b) Ta có: \(A=\dfrac{10^{2009}+1}{10^{2010}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{2010}+10}{10^{2010}+1}=1+\dfrac{9}{10^{2010}+1}\)
Ta có: \(B=\dfrac{10^{2010}+1}{10^{2011}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{2011}+10}{10^{2011}+1}=1+\dfrac{9}{10^{2011}+1}\)
Ta có: \(10^{2010}+1< 10^{2011}+1\)
\(\Leftrightarrow\dfrac{9}{10^{2010}+1}>\dfrac{9}{10^{2011}+1}\)
\(\Leftrightarrow\dfrac{9}{10^{2010}+1}+1>\dfrac{9}{10^{2011}+1}+1\)
\(\Leftrightarrow10A>10B\)
hay A>B
17 tháng 1 2023
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
10 tháng 10 2023
a) \(3\cdot24^{10}=3\cdot6^{10}\cdot4^{10}=3\cdot3^{10}\cdot2^{10}\cdot2^{20}\)
\(=3^{11}\cdot2^{30}\)
\(4^{30}=2^{30}\cdot2^{30}=2^{30}\cdot4^{15}\)
Ta có \(4^{15}>3^{15}>3^{11}\) nên \(4^{15}>3^{11}\)
Khi đó \(4^{15}\cdot2^{30}>3^{11}\cdot2^{30}\) hay \(4^{30}>3\cdot24^{10}\)
b) \(\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{19}{9^2\cdot10^2}\)
\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+...+\dfrac{19}{81\cdot100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
Vậy dãy trên nhỏ hơn 1
NN
10 tháng 10 2023
a/
\(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}.2^{30}=\left(2^2\right)^{15}.2^{30}=4^{15}.2^{30}\)
\(3.24^{10}=3.3^{10}.\left(2^3\right)^{10}=3^{11}.2^{30}< 3^{15}.2^{30}\)
\(\Rightarrow4^{30}=4^{15}.2^{30}>3^{15}.2^{30}>3^{11}.2^{30}=3.24^{10}\)
b/
\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}=\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}=\)
\(=1-\dfrac{1}{10^2}< 1\)
11 tháng 9 2023
\(\dfrac{47}{95}\) và \(\dfrac{35}{69}\)
\(\dfrac{47}{95}< \dfrac{1}{2}\) và \(\dfrac{35}{69}>\dfrac{1}{2}\)
Vậy \(\dfrac{47}{95}< \dfrac{35}{69}\)
\(\dfrac{53}{103}\) và \(\dfrac{71}{145}\)
\(\dfrac{53}{103}>\dfrac{1}{2}\) và \(\dfrac{71}{145}< \dfrac{1}{2}\)
Vậy \(\dfrac{53}{103}>\dfrac{71}{145}\)
\(\dfrac{2009}{2010}\) và \(\dfrac{2005}{2006}\)
\(1-\dfrac{2009}{2010}=\dfrac{1}{2010}\) và \(1-\dfrac{2005}{2006}=\dfrac{1}{2006}\)
Vậy \(\dfrac{2009}{2010}>\dfrac{2005}{2006}\)
\(\dfrac{783}{901}\) và \(\dfrac{738}{915}\)
\(\dfrac{738}{915}< \dfrac{783}{915}< \dfrac{783}{901}\)
Vậy \(\dfrac{783}{901}>\dfrac{738}{915}\)
27 tháng 1 2021
Ta có :
\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)
\(\Leftrightarrow A>B\)
17 tháng 5 2022
\(Q=\dfrac{2010+2011+2012}{2011+2012+2013}=\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)
Ta có: \(\dfrac{2010}{2011+2012+2013}< \dfrac{2010}{2011}\)
\(\dfrac{2011}{2011+2012+2013}< \dfrac{2011}{2012}\)
\(\dfrac{2012}{2011< 2012< 2013}< \dfrac{2012}{2013}\)
\(\Rightarrow\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)
\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(P>Q\)
NM
0
HQ
Hà Quang Minh
Giáo viên
27 tháng 11 2023
a) HS tự thực hiện
b) $\frac{5}{6}$ < 1 ; $\frac{3}{2} > 1$
$\frac{9}{{19}}$ < 1 ; $\frac{7}{7}$ = 1
$\frac{{49}}{{46}}$ > 1 ; $\frac{{32}}{{71}}$ < 1
c) Ba phân số bé hơn 1 là: $\frac{2}{7};\,\,\,\frac{{11}}{{25}};\,\,\,\frac{{37}}{{59}}$
Ba phân số lớn hơn 1 là: $\frac{7}{2};\,\,\,\frac{{15}}{7};\,\,\,\,\frac{{33}}{{12}}$
Ba phân số bằng 1 là: $\frac{9}{9};\,\,\,\,\frac{{25}}{{25}};\,\,\,\,\frac{{47}}{{47}}$
\(\dfrac{1}{2010}>0\\ \dfrac{-7}{19}< 0\\ \Rightarrow\dfrac{-7}{19}< \dfrac{1}{2010}\)
`1/2010>0`
`-7/19<0`
`=>1/2010> -7/19`