So sánh A và B
A=\(\frac{13^{15}+1}{13^{14}+1}\)
B=\(\frac{13^{14}+1}{13^{13}+1}\)
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NA
0
TT
1
PT
1
30 tháng 8 2017
Vì 1113 . 1115 = 1114 . 1114 = 1128 nên \(\frac{11^{13}+1}{11^{14}+1}=\frac{11^{14}+1}{11^{15}+1}\)
HD
5
TK
8 tháng 6 2018
Đặt : \(A=\frac{2018^{13}+1}{2018^{14}+1}\); \(B=\frac{2018^{2012}+1}{2018^{2013}+1}\)
Ta có :
\(2018A=\frac{2018.\left(2018^{13}+1\right)}{2018^{14}+1}\)
\(2018A=\frac{2018^{14}+2018}{2018^{14}+1}=\frac{2018^{14}+1+2017}{2018^{14}+1}=\frac{2018^{2014}+1}{2018^{14}+1}+\frac{2017}{2018^{14}+1}=1+\frac{2017}{2018^{14}+1}\)
\(2018B=\frac{2018.\left(2018^{12}+1\right)}{2018^{13}+1}\)
\(2018B=\frac{2018^{13}+2018}{2018^{13}+1}=\frac{2018^{13}+1+2017}{2018^{13}+1}=\frac{2018^{13}+1}{2018^{13}+1}+\frac{2017}{2018^{13}+1}=1+\frac{2017}{2018^{13}+1}\)
Vì 201814 + 1 > 201813 + 1 nên \(\frac{2017}{2018^{14}+1}< \frac{2017}{2018^{13}+1}\)
\(\Rightarrow1+\frac{2017}{2018^{14}+1}< 1+\frac{2017}{2018^{13}+1}\)Hay : A < B
Vậy A < B
IA
8 tháng 6 2018
Đặt \(A=\frac{2018^{13}+1}{2018^{14}+1}\)và \(B=\frac{2018^{12}+1}{2018^{13}+1}\)
Ta có :
\(2018A=\frac{\left(2018^{13}+1\right)\times2018}{2018^{14}+1}\) \(2018B=\frac{\left(2018^{12}+1\right)\times2018}{2018^{13}+1}\)
\(2018A=\frac{2018^{14}+2018}{2018^{14}+1}\) \(2018B=\frac{2018^{13}+2018}{2018^{13}+1}\)
\(2018A=\frac{2018^{14}+1+2017}{2018^{14}+1}\) \(2018B=\frac{2018^{13}+1+2017}{2018^{13}+1}\)
\(2018A=1+\frac{2017}{2018^{14}+1}\) \(2018B=1+\frac{2017}{2018^{13}+1}\)
Vì \(\frac{2017}{2018^{14}+1}< \frac{2017}{2018^{13}+1}\)
\(\Rightarrow2018A< 2018B\)
\(\Rightarrow A< B\)
Vậy : \(\frac{2018^{13}+1}{2018^{14}+1}< \frac{2018^{12}+1}{2018^{13}+1}\)
14 tháng 3 2020
Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
14 tháng 3 2020
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
DT
2
29 tháng 10 2016
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=\frac{13^{16}+1+12}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=\frac{13^{17}+1+12}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Ta thấy:
\(13^{16}+1< 13^{17}+1\)
\(\Rightarrow\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
hay \(A>B\)
Vậy \(A>B.\)
DT
2
Ta có: A<1
B> 1
=> A < B
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