Cho A= \(\frac{2017^{2017}+2}{2017^{2017}-1}\) và B=\(\frac{2017^{2017}}{2017^{2017}-3}\). Hãy so sánh A và B.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Các câu dễ bạn tự làm nha:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2017^{2017}+1}{2017^{2018}+1}< 1\)
\(A< \dfrac{2017^{2017}+1+2016}{2017^{2018}+1+2016}\Rightarrow A< \dfrac{2017^{2017}+2017}{2017^{2018}+2017}\Rightarrow A< \dfrac{2017\left(2017^{2016}+1\right)}{2017\left(2017^{2017}+1\right)}\Rightarrow A< \dfrac{2017^{2016}+1}{2017^{2017}+1}=B\)\(A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\frac{2017^{2017}+2}{2017^{2017}-1}\)=\(\frac{\left(2017^{2017}-1\right)+3}{2017^{2017}-1}\)=\(1\)+\(\frac{3}{2017^{2017}-1}\)
B=\(\frac{2017^{2017}}{2017^{2017}-3}\)=\(\frac{\left(2017^{2017}-3\right)+3}{2017^{2017}-3}\)=\(1\)+\(\frac{3}{2017^{2017}-3}\)
Vì \(2017^{2017}-1\)\(>\)\(2017^{2017}-3\)nên \(\frac{3}{2017^{2017}-1}\)\(< \)\(\frac{3}{2017^{2017}-3}\)=> A<B
vậy A<B
chúc bạn học giỏi
k giùm mk nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\frac{2017^{2017+2}}{2017^{2017-1}}\) B =\(\frac{2017^{2017}}{2017^{2017-3}}\)
=\(\frac{2017^{2019}}{2017^{2016}}\) =\(\frac{2017^{2017}}{2017^{2014}}\)
=\(\frac{2017^{2016}\cdot2017^3}{2017^{2016}}\)=\(2017^3\) =\(\frac{2017^{2014}\cdot2017^3}{2017^{2014}}\)=\(2017^3\)
Vì\(2017^3=2017^3\) nên A=B
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu a : Cộng 2 vế cho 6 ta được :
\(7+6......7+\sqrt{37}\)
Mà : \(6=\sqrt{36}< \sqrt{37}\)
\(\Rightarrow7+6< \sqrt{37}+1\)
\(\Rightarrow7< \sqrt{37}+1\)
Cách khác của câu a.
Ta có : \(\sqrt{37}>\sqrt{36}=6\)
\(\Rightarrow\sqrt{37}+1>6+1=7\)
Vậy \(\sqrt{37}+1>7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\frac{2+2^2+...+2^{2017}}{1-2^{2017}}\)
Đặt \(B=2+2^2+...+2^{2017}\)
\(\Rightarrow2B=2^2+2^3+...+2^{2018}\)
\(\Rightarrow2B-B=\left(2^2+2^3+...+2^{2018}\right)-\left(2+...+2^{2017}\right)\)
\(\Rightarrow B=2^{2018}-2\)
\(\Rightarrow A=\frac{2^{2018}-2}{1-2^{2017}}\)
\(\Rightarrow A=\frac{-2.\left(1-2^{2017}\right)}{1-2^{2017}}\)
\(\Rightarrow A=-2\)
b)Đề phải là CM: \(A< \frac{2017}{2016^2}\)
\(A=\frac{1}{2017}+\frac{2}{2017^2}+...+\frac{22017}{2017^{2017}}+\frac{2018}{2017^{2018}}\)
\(\Rightarrow2017A=1+\frac{2}{2017}+...+\frac{22017}{2017^{2016}}+\frac{2018}{2017^{2017}}\)
\(\Rightarrow2017A-A=\left(1+...+\frac{2018}{2017^{2017}}\right)-\left(\frac{1}{2017}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\right)\)
\(\Rightarrow2016A=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}-\frac{2018}{2017^{2018}}\)
Đặt \(\Rightarrow S=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}\)
\(\Rightarrow2017S=2017+1+\frac{1}{2017}+...+\frac{1}{2017^{2016}}\)
\(\Rightarrow2017S-S=\left(2017+1+...+\frac{1}{2017^{2016}}\right)-\left(1+...+\frac{1}{2017^{2017}}\right)\)
\(\Rightarrow2016S=2017-\frac{1}{2017^{2017}}< 2017\)
\(\Rightarrow2016S< 2017\)
\(\Rightarrow S< \frac{2017}{2016}\)
\(\Rightarrow2016A< \frac{2017}{2016}\)
\(\Rightarrow A< \frac{2017}{2016^2}\left(đpcm\right)\)
Ta có : \(A=\frac{2017^{2017}+2}{2017^{2017}-1}=\frac{2017^{2017}-1+3}{2017^{2017}-1}=1+\frac{3}{2017^{2017}-1}\)
Lại có : \(B=\frac{2017^{2017}}{2017^{2017}-3}=\frac{2017^{2017}-3+3}{2017^{2017}-3}=1+\frac{3}{2017^{2017}-3}\)
Nhận thấy : 20172017 - 1 > 20172017 - 3
=> \(\frac{3}{2017^{2017}-1}< \frac{3}{2017^{2017}-3}\)
=> \(1+\frac{3}{2017^{2017}-1}< 1+\frac{3}{2017^{2017}-3}\)
=> A < B