So sánh A=\(\frac{10^{2011}+1}{10^{2012}+1}\)và B=\(\frac{10^{2012}+1}{10^{2013}+1}\)
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N
1
ML
5 tháng 7 2017
a) \(\frac{2^{10}+1}{2^{10}-1}\)và \(\frac{2^{10}-1}{2^{10}-3}\)
Ta có chính chất phân số trung gian là \(\frac{2^{10}+1}{2^{10}-3}\)
\(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}\) ; \(\frac{2^{10}-1}{2^{10}-3}< \frac{2^{10}+1}{2^{10}-3}\)
Vì \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}>\frac{2^{10}-1}{2^{10}-3}\)
Nên \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}-1}{2^{10}-3}\)
b) \(A=\frac{2011}{2012}+\frac{2012}{2013}\)và \(B=\frac{2011+2012}{2012+2013}\)
Ta có : \(A=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}=B\)
Vậy A > B
Có gì sai cho sorry
5 tháng 7 2017
a,
\(\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}=\frac{2^{10}-1}{2^{10}-3}\)
b,
\(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
KT
1
1 tháng 7 2016
Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
\(=>B=\frac{10^{2012}+1}{10^{2013}+1}< \frac{10^{2012}+1+9}{10^{2013}+1+9}\)
\(< \frac{10^{2012}+10}{10^{2013}+10}\)
\(< \frac{10.\left(10^{2011}+1\right)}{10.\left(10^{2012}+1\right)}\)
\(< \frac{10^{2011}+1}{10^{2012}+1}=A\)
=> B < A
Ủng hộ mk nha ^_-
YS
28 tháng 2 2016
So sánh 2 phân số sau $\frac{10^{2011}+10}{10^{2012}+10}v\text{à}\frac{10^{2012}-10}{10^{2013}-10}$102011+10102012+10 và102012−10102013−10
kick dzô chữ xanh là được!! OK
28 tháng 2 2016
Ta có :
10. A = \(\frac{10.\left(10^{2011}+1\right)}{10^{2012}+1}\)
= \(\frac{10^{2012}+10}{10^{2012}+1}\)
= \(\frac{10^{2012}+1+9}{10^{2012}+1}\)
= \(\frac{10^{2012}+1}{10^{2012}+1}-\frac{9}{10^{2012}+1}\)
= 1 - \(\frac{9}{10^{2012}+1}\)
10 . B = \(\frac{10.\left(10^{2012}+1\right)}{10^{2013}+1}\)
= \(\frac{10^{2013}+10}{10^{2013}+1}\)
= \(\frac{10^{2013}+1+9}{10^{2013}+1}\)
= 1 - \(\frac{9}{10^{2013}+1}\)
Vì \(\frac{9}{10^{2012}+1}\) >\(\frac{9}{10^{2013}+1}\) nên 10.A > 10.B
=> A >B
Vậy ...........
I2
1
MN
21 tháng 3 2020
Có : \(A=\frac{10^{2012}-10}{10^{2013}-10}\)
\(\Leftrightarrow10A=\frac{10^{2013}-100}{10^{2013}-10}\)
\(\Leftrightarrow10A=\frac{10^{2013}-10-90}{10^{2013}-10}\)
\(\Leftrightarrow10A=1-\frac{90}{10^{2013}-10}\)
Có : \(B=\frac{10^{2011}+10}{10^{2012}+10}\)
\(\Leftrightarrow10B=\frac{10^{2012}+100}{10^{2012}+10}\)
\(\Leftrightarrow10B=\frac{10^{2012}+10+90}{10^{2012}+10}\)
\(\Leftrightarrow B=1+\frac{90}{10^{2012}+10}\)
Ta thấy : \(1-\frac{90}{10^{2013}-10}< 1\)
\(1+\frac{90}{10^{2012}+10}>1\)
\(\Leftrightarrow1-\frac{90}{10^{2013}-10}< 1+\frac{90}{10^{2012}+10}\)
\(\Leftrightarrow A< B\)
PT
2
S
2 tháng 3 2017
Vì \(\frac{10^{2011}+1}{10^{2012}+1}< 1\)
=> \(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+1+9}{10^{2012}+1+9}=\frac{10^{2011}+10}{10^{2012}+10}=\frac{10\left(10^{2010}+1\right)}{10\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)
Vậy A > B
12 tháng 1 2019
b,Ta có
\(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
\(\Rightarrow P>Q\)
12 tháng 1 2019
\(A=\frac{-10}{20}+\frac{-10}{30}+\frac{-10}{42}+\frac{-10}{56}+\frac{-10}{72}+\frac{-10}{90}+\frac{-10}{110}\)
\(=-10\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)\)
\(=-10\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=-10\left(\frac{1}{4}-\frac{1}{11}\right)\)
\(=\frac{-35}{22}\)
15 tháng 3 2019
a)
\(10A=\frac{10^{2002}+10}{10^{2002}+1}=1+\frac{9}{10^{2002}+1}\)
\(10B=\frac{10^{2003}+10}{10^{2003}+1}=1+\frac{9}{10^{2003}+1}\)
=> 10A > 10B => A > B
\(B< \frac{10^{2012}+1+9}{10^{2013}+1+9}=\frac{10^{2012}+10}{10^{2013}+10}=\frac{10\left(10^{2011}+1\right)}{10\left(10^{2012}+1\right)}=\frac{10^{2011}+1}{10^{2012}+1}=A\)
Vậy A > B
Áp dụng bất đẳng thức :
\(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\)
Ta có :
\(B=\frac{10^{2012}+1}{10^{2013}+1}< \frac{10^{2012}+1+9}{10^{2013}+1+9}=\frac{10^{2012}+10}{10^{2013}+10}=\frac{10\left(10^{2011}+1\right)}{10\left(10^{2012}+1\right)}=\frac{10^{2011}+1}{10^{2012}+1}=A\)
\(\Leftrightarrow B< A\)