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1: so sánh 2016/2017+2017/2018
vì 2016/2017 > 1/2017 >1/2018 =
> 2016/2017+2017/2018 >1/2018+2017/2018=1
vậy .....
Có \(a\left(b+1\right)< b\left(a+1\right)\Leftrightarrow ab+a< ab+b\)
\(\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\)
Áp dụng \(\frac{2^{2018}}{3^{2019}}< \frac{2^{2018}+1}{3^{2019}+1}\)
Ta có:
\(1-\frac{a}{b}=\frac{b-a}{b}\)
\(1-\frac{a+1}{b+1}=\frac{b+1-a-1}{b+1}=\frac{b-a}{b+1}\)
Vì b < b + 1 và a < b; a, b nguyên dương => b - a > 0 nên \(\frac{b-a}{b}>\frac{b-a}{b+1}\)
Do đó \(1-\frac{a}{b}>1-\frac{a+1}{b+1}\)
\(\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\)
Áp dụng chứng minh tương tự nhé bạn
Đặt \(A=\frac{2^{2017}+1}{2^{2018}+1}\Rightarrow2A=\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
\(B=\frac{2^{2018}+1}{2^{2019}+1}\Rightarrow2B=\frac{2^{2019}+2}{2^{2019}+1}=\frac{2^{2019}+1+1}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Vì \(2^{2019}+1>2^{2018}+1\Rightarrow\frac{1}{2^{2019}+1}< \frac{1}{2^{2018}+1}\)
\(\Rightarrow2A>2B\Rightarrow A>B\)
Nhận xét :
\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
....
\(\frac{1}{2018^2}=\frac{1}{2018.2018}< \frac{1}{2017.2018}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
=> \(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
=> \(A< 1-\frac{1}{2018}< 1=B\)
Vậy \(A< B\)
A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2018.2018}\)
< \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
= \(1-\frac{1}{2018}=\frac{2017}{2018}< 1\)
=> A < B
Vậy A < B