So Sánh \(\frac{2017^{99}+1}{2017^{100}+1}\)và \(\frac{2017^{100}+1}{2017^{101}+1}\)
Giải giùm mình với
K
Khách
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NC
3
26 tháng 5 2017
Ta có: \(A=\frac{2017^{99}+1}{2017^{100}+1}\Rightarrow2017A=\frac{2017^{100}+2017}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)
\(B=\frac{2017^{100}+1}{2017^{101}+1}\Rightarrow2017B=\frac{2017^{101}+2017}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)
\(\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)
\(\Rightarrow2017A>2017B\Rightarrow A>B\)
Vậy...
26 tháng 5 2017
Đặt \(A=\frac{2017^{99}+1}{2017^{100}+1}\)nên \(2017A=\frac{2017^{100}+2017}{2017^{100}+1}=\frac{2017^{100}+1+2016}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)
\(B=\frac{2017^{100}+1}{2017^{101}+1}\)nên \(2017B=\frac{2017^{101}+2017}{2017^{101}+1}=\frac{2017^{101}+1+2016}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)
Vì \(1=1;\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)
Hay \(2017A>2017B\)nên \(A>B\)
Vây \(\frac{2017^{99}+1}{2017^{1001}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)
PD
0
BT
2 tháng 6 2017
\(A=\frac{2017^{99}}{2017^{100}-2}\)
=> \(2017A=\frac{2017^{100}}{2017^{100}-2}=\frac{2017^{100}-2+2}{2017^{100}-2}=1+\frac{2}{2017^{100}-2}\)
\(B=\frac{2017^{100}}{2017^{101}-2}\)
=>\(2017B=\frac{2017^{101}}{2017^{101}-2}=\frac{2017^{101}-2+2}{2017^{101}-2}=1+\frac{2}{2017^{101}-2}\)
Do \(\frac{2}{2017^{100}-2}>\frac{2}{2017^{101}-2}\)
Nên 2017A > 2017B
Vậy A > B
T
13 tháng 2 2018
Ta có: \(A=\frac{2017^{100}}{1+2017+2017^2+2017^3+...+2017^{100}}\)
\(\Leftrightarrow A=\frac{\left[\left(20.100\right)+16+1\right]^{100}}{1+2017+2017^2+2017^3+...+2017^{10}}\)
\(B=\frac{2016^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)
\(\Leftrightarrow B=\frac{\left[\left(20.100+16\right)\right]^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)
Ta có hai tổng A và B mới để so sánh:
\(A=\frac{\left[\left(20.100\right)+16+1\right]^{100}}{1+2017+2017^2+2017^3+...+2017^{100}}\)
\(B=\frac{\left[\left(20.100\right)+16\right]^{100}}{1+2016+2016^2+2016^3+...+2016^{100}}\)
Tới đây đơn giản rồi. Bạn làm tiếp đi nhé! Mẹ mình bắt tắt máy không cho làm nên đành dừng lại ở đây thôi! Thông cảm :V
24 tháng 4 2017
a/ Ta có
\(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)
\(=1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)
\(=1+2\left(\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)\)
\(=2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
Thế lại bài toán ta được:
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)
\(=\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\)
24 tháng 4 2017
b/ Ta có:
A - B\(=\frac{-21}{10^{2016}}+\frac{12}{10^{2016}}+\frac{21}{10^{2017}}-\frac{12}{10^{2017}}\)
\(=\frac{9}{10^{2017}}-\frac{9}{10^{2016}}< 0\)
Vậy A < B
3 tháng 5 2019
\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100\cdot\left[100^{2017}+1\right]}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100\cdot\left[100^{2018}+1\right]}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)
Tự so sánh
KN
3 tháng 5 2019
\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1}{100^{2018}+1}+\frac{99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)(1)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1}{100^{2019}+1}+\frac{99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)(2)
Từ (1) và (2) suy ra 100A > 100B hay A > B
vì 2017100 + 1 < 2017101 + 1
\(\Rightarrow\frac{2017^{100}+1}{2017^{101}+1}< \frac{2017^{100}+1+2016}{2017^{101}+1+2016}=\frac{2017^{100}+2017}{2017^{101}+2017}=\frac{2017.\left(2017^{99+1}\right)}{2017.\left(2017^{100}+1\right)}=\frac{2017^{99}+1}{2017^{100}+1}\)
Vậy \(\frac{2017^{99}+1}{2017^{100}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)
so sánh 2 phân số cùng mẫu thì ta xét tử
đừng nói không làm được chứ