So Sánh
A=\(\frac{10^{2013}+1}{10^{2014}+1}\)
B=\(\frac{10^{2014}+1}{10^{2015}+1}\)
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có :
\(B=\frac{10^{2015}+1}{10^{2014}+1}>1\)
\(\Rightarrow\frac{10^{2015}+1}{10^{2014}+1}>\frac{10^{2015}+1+9}{10^{2014}+1+9}\) \(=\frac{10^{2015}+10}{10^{2014}+10}=\frac{10.\left(10^{2014}+1\right)}{10.\left(10^{2013}+1\right)}\)
\(=\frac{10^{2014}+1}{10^{2013}+1}=A\)
\(\Rightarrow B>A\)
Vậy B > A
k cho mk nhé
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{10^{2014}+1}{10^{2015}+1}< \frac{10^{2014}+1+9}{10^{2015}+1+9}=\frac{10^{2014}+10}{10^{2015}+10}=\frac{10\left(10^{2013}+1\right)}{10\left(10^{2014}+1\right)}=\frac{10^{2013}+1}{10^{2014}+1}=A\)
\(\Rightarrow\)\(B< A\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
áp dụng tính chất
nếu a/b>1thì a/b<(a+n)/(b+n)
=)))))))))))))))))
2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)
\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)
Nên suy ra \(10A>10B\Rightarrow A>B\)
\(A=\frac{10^{2015}-1}{10^{2016}^{ }-1}=\frac{10^{2015}}{10^{2016}}=\frac{1}{1},B=\frac{10^{2014}-1}{10^{2015}-1}=\frac{10^{2014}}{10^{2015}}=\frac{1}{1}A=B\Rightarrow\)
\(A=\frac{10^{2012}+1}{10^{2013}+1}\)
\(10A=\frac{10\cdot\left[10^{2012}+1\right]}{10^{2013}+1}=\frac{10^{2013}+10}{10^{2013}+1}=\frac{10^{2013}+1+9}{10^{2013}+1}=1+\frac{9}{10^{2013}+1}\)
\(B=\frac{10^{2013}+1}{10^{2014}+1}\)
\(10B=\frac{10\cdot\left[10^{2013}+1\right]}{10^{2014}+1}=\frac{10^{2014}+10}{10^{2014}+1}=\frac{10^{2014}+1+9}{10^{2014}+1}=1+\frac{9}{10^{2014}+1}\)
Mà \(1+\frac{9}{10^{2013}+1}>1+\frac{9}{10^{2014}+1}\)
Nên \(10A>10B\)
Hay \(A>B\)
Vậy : A > B
Vì \(\frac{10^{2014}+1}{10^{2015}+1}< 1\Rightarrow B=\frac{10^{2014}+1}{10^{2015}+1}< \frac{10^{2014}+1+9}{10^{2015}+1+9}\)
\(\Rightarrow B< \frac{10^{2014}+10}{10^{2015}+10}\)
\(\Rightarrow B< \frac{10\left(10^{2013}+1\right)}{10\left(10^{2014}+1\right)}\)
\(\Rightarrow B< \frac{10^{2013}+1}{10^{2014}+1}\)
\(\Rightarrow B< A\)
Vậy A > B
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