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1 tháng 4 2017
a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)
Vì \(10^{16}+10^{15}>10^{16}+10\)
\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)
Hay A>B
b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)
\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)
Vì \(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)
\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)
Hay C > D
3 tháng 5
a, Ta có : \(1 0^{15} \cdot 11 = 1 0^{15} \left(\right. 10 + 1 \left.\right) = 1 0^{16} + 1 0^{15}\)
Vì \(1 0^{16} + 1 0^{15} > 1 0^{16} + 10\)
\(\Rightarrow \frac{1 0^{16} + 1 0^{15}}{1 0^{16} + 1} > \frac{1 0^{16} + 10}{1 0^{16} + 1}\)
Hay A>B
b, Ta có : \(C = \frac{1 0^{10} + 1}{1 0^{10} - 1} = \frac{1 0^{10}}{1 0^{10} - 1} + \frac{1}{1 0^{10} - 1}\)
\(D = \frac{1 0^{10} - 1}{1 0^{13} - 3} = \frac{1 0^{10}}{1 0^{13} - 3} + \frac{- 1}{1 0^{13} - 3}\)
Vì \(\frac{1 0^{10}}{1 0^{10} - 1} > \frac{1 0^{10}}{1 0^{13} - 3} ; \frac{1}{1 0^{10} - 1} > \frac{- 1}{1 0^{13} - 3}\)
\(\Rightarrow \frac{1 0^{10} + 1}{1 0^{10} - 1} > \frac{1 0^{10} - 1}{1 0^{13} - 3}\)
Hay C > D
30 tháng 3 2017
Ta có :
\(A=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+1+1}{10^{10}-1}=\dfrac{\left(10^{10}-1\right)+1}{10^{10}-1}=1+\dfrac{2}{2016^{10}-1}\) \(\left(1\right)\)
\(B=\dfrac{10^{10}-1}{10^{10}-3}=\dfrac{10^{10}-3-1+3}{10^{10}-3}=\dfrac{\left(10^{10}-3\right)+2}{10^{10}-3}=1+\dfrac{2}{10^{10}-3}\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\) \(\Rightarrow\) \(A< B\)
Chúc bn học tốt!!
30 tháng 3 2017
Ta có A=\(\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+2}{10^{10}-1}=\dfrac{10^{10}-1}{10^{10}-1}+\dfrac{2}{10^{10}-1}\)
\(=1+\dfrac{2}{10^{10}-1}\)
B=\(\dfrac{10^{10}-1}{10^{10}-3}=\)\(\dfrac{10^{10}-3+2}{10^{10}-3}\)=\(\dfrac{10^{10}-3}{10^{10}-3}+\dfrac{2}{10^{10}-3}\)
=\(1+\dfrac{2}{10^{10}-3}\)
Vì \(\dfrac{2}{10^{10}-1}>\dfrac{2}{10^{10}-3}\)
\(\Rightarrow\) A>B
Chúc bạn học tốt tick mik nha
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 : \(1-\frac{90}{10^{2013}-10}< 1\)
\(1+\frac{90}{10^{2012}+10}>1\)
\(\Leftrightarrow1-\frac{90}{10^{2013}-10}< 1+\frac{90}{10^{2012}+10}\)
\(\Leftrightarrow A< B\)
TT
3
14 tháng 11 2019
s=1+10+10^2+.....+10^10
10s=10+10^2+10^3+.......+10^11
=))10s-s=10^11-1
=)9s=10^11-1
s=(10^11-1):9
mk làm hơi tắt bước , xl nhé:>>>>
23 tháng 2 2017
a) Ta có: \(10A=\frac{10^{16}+10}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
\(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
Vậy A > B
b) Ta có: \(\frac{1}{10}C=\frac{10^{1992}+1}{10^{1992}+10}=1+\frac{10^{1992}+1}{9}\)
\(\frac{1}{10}D=\frac{10^{1993}+1}{10^{1993}+10}=1+\frac{10^{1993}+1}{9}\)
\(\frac{10^{1992}+1}{9}< \frac{10^{1993}+1}{9}\Rightarrow1+\frac{10^{1992}+1}{9}< 1+\frac{10^{1993}+1}{9}\)
\(\Rightarrow\frac{1}{10}C< \frac{1}{10}D\)
\(\Rightarrow C< D\)
Vậy C < D
L
11 tháng 8 2019
Bài giải
Ta có :
\(\frac{13}{14}=1-\frac{1}{14}\)
\(\frac{12}{13}=1-\frac{1}{13}\)
Vì \(\frac{1}{14}< \frac{1}{13}\) \(\Rightarrow\text{ }\frac{13}{14}>\frac{12}{13}\)
L
11 tháng 8 2019
b, Bài giải
\(A=\frac{10^{10}+5}{10^{10}-1}=\frac{10^{10}-1+6}{10^{10}-1}=\frac{10^{10}-1}{10^{10}-1}+\frac{6}{10^{10}-1}=1+\frac{6}{10^{10}-1}\)
\(B=\frac{10^{10}+4}{10^{10}-2}=\frac{10^{10}-2+6}{10^{10}-2}=\frac{10^{10}-2}{10^{10}-2}+\frac{6}{10^{10}-2}=1+\frac{6}{10^{10}-2}\)
Vì \(\frac{6}{10^{10}-1}>\frac{6}{10^{10}-2}\) \(\Rightarrow\text{ }\frac{10^{10}+5}{10^{10}-1}>\frac{10^{10}+4}{10^{10}-2}\)
\(\Rightarrow\text{ }A>B\)
XO
24 tháng 7 2020
a) Ta có : 10A = \(\frac{10\left(10^{2004}+1\right)}{10^{2005}+1}=\frac{10^{2005}+10}{10^{2005}+1}=1+\frac{9}{10^{2005}+1}\)
Lại có 10B = \(\frac{10\left(10^{2005}+1\right)}{10^{2006}+1}=\frac{10^{2006}+10}{10^{2006}+1}=1+\frac{9}{10^{2006}+1}\)
Vì \(\frac{9}{10^{2005}+1}>\frac{9}{10^{2006}+1}\Rightarrow1+\frac{9}{10^{2005}+1}>1+\frac{9}{10^{2006}+1}\)
=> 10A > 10B
=> A > B
b) Ta có A = \(\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
Lại có B = \(\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1-\frac{2}{20^{10}-3}\)
=> A < B
21 tháng 4 2017
Ta có: \(A=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+2}{10^{10}-1}=1+\dfrac{2}{10^{10}-1}\)
\(B=\dfrac{10^{10}-1}{10^{10}-3}=\dfrac{10^{10}-3+2}{10^{10}-3}=1+\dfrac{2}{10^{10}-3}\)
Vì \(\dfrac{2}{10^{10}-1}< \dfrac{2}{10^{10}-3}\Rightarrow1+\dfrac{2}{10^{10}-1}< 1+\dfrac{2}{10^{10}-3}\)
\(\Rightarrow A< B\)
Vậy A < B
AT
21 tháng 4 2017
\(A=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+2}{10^{10}-1}=1+\dfrac{2}{10^{10}-1}\)
\(B=\dfrac{10^{10}-1}{10^{10}-3}=\dfrac{10^{10}-3+2}{10^{10}-3}=1+\dfrac{2}{10^{10}-3}\)
Vì \(10^{10}-1>10^{10}-3\) nên ta có
\(\dfrac{2}{10^{10}-1}< \dfrac{2}{10^{10}-3}\)
Vậy \(A< B\)
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