Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu dễ bạn tự làm nha:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2017^{2017}+1}{2017^{2018}+1}< 1\)
\(A< \dfrac{2017^{2017}+1+2016}{2017^{2018}+1+2016}\Rightarrow A< \dfrac{2017^{2017}+2017}{2017^{2018}+2017}\Rightarrow A< \dfrac{2017\left(2017^{2016}+1\right)}{2017\left(2017^{2017}+1\right)}\Rightarrow A< \dfrac{2017^{2016}+1}{2017^{2017}+1}=B\)\(A< B\)
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
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
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 \(\Rightarrow S=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}\)
\(\Rightarrow2017S=2017+1+\frac{1}{2017}+...+\frac{1}{2017^{2016}}\)
\(\Rightarrow2017S-S=\left(2017+1+...+\frac{1}{2017^{2016}}\right)-\left(1+...+\frac{1}{2017^{2017}}\right)\)
\(\Rightarrow2016S=2017-\frac{1}{2017^{2017}}< 2017\)
\(\Rightarrow2016S< 2017\)
\(\Rightarrow S< \frac{2017}{2016}\)
\(\Rightarrow2016A< \frac{2017}{2016}\)
\(\Rightarrow A< \frac{2017}{2016^2}\left(đpcm\right)\)
a,xét A:
A=2016x2018
A=2016x(2017+1)
A=2016x1+2016x2017
Xét B:
B=2017x2017
B=2017x(2016+1)
B=2017x2016+2017x1
Vì cả 2 vế Avà B đều có 2017x2016maf 2017x1>2016x1 nên A bé hơn B
Ta có :
\(\frac{2015}{2016}>\frac{2015}{2016+2017}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>B\)
Chúc bạn học tốt !!!
\(A=\frac{2015}{2016}+\frac{2016}{2017}\) \(B=\frac{2015+2016}{4033}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}\) \(B=\frac{2015}{4033}+\frac{2016}{4033}\)
\(\Rightarrow A>B\)
Đặt C = 1 + 2017 + 20172 + ... + 20172016 ; D = 1 + 2016 + 20162 + ... + 20162016
Ta có : 2017C = 2017 + 20172 + 20173 + ... + 20172017
=> 2016C = 2017C - C = 20172017 - 1\(\Rightarrow C=\frac{2017^{2017}-1}{2016}\)
2016D = 2016 + 20162 + 20163 + ... + 20162017
=> 2015D = 2016D - D = 20162017 - 1\(\Rightarrow D=\frac{2016^{2017}-1}{2015}\)
\(\Rightarrow A=\frac{2017^{2017}}{\frac{2017^{2017}-1}{2016}}=\frac{2017^{2017}.2016}{2017^{2017}-1}\);\(B=\frac{2016^{2017}}{\frac{2016^{2017}-1}{2015}}=\frac{2016^{2017}.2015}{2016^{2017}-1}\)
Ta có : 20172017.2016.(20162017 - 1) - 20162017.2015.(20172017 - 1)
= 20172017.20162017.2016 - 20172017.2016 - 20172017.20162017.2015 + 20162017.2015
= 20172017.20162017 - 20172017.2016 + 20162017.2015
= 20172017.(20162017 - 2016) + 20162017.2015 > 0
=> A > B
Ta có
\(A=1:\frac{1+2017+2017^2+...+2017^{2016}}{2017^{2017}}\)
\(B=1:\frac{1+2016+2016^2+...2016^{2016}}{2016^{2017}}\)
\(A=1:\left(\frac{1}{2017^{2017}}+\frac{1}{2017^{2016}}+\frac{1}{2017^{2015}}+...+\frac{1}{2017}\right)\)
\(B=1:\left(\frac{1}{2016^{2017}}+\frac{1}{2016^{2016}}+\frac{1}{2016^{2015}}+...+\frac{1}{2016}\right)\)
Có 20172017>20162017 ; 20172016>20162016 ; 20172015>20162015;..... ; 2017>2016
=> \(\frac{1}{2017^{2017}}< \frac{1}{2016^{2017}};\frac{1}{2017^{2016}}< \frac{1}{2016^{2016}};\frac{1}{2017^{2015}}< \frac{1}{2016^{2015}};...;\frac{1}{2017}< \frac{1}{2016}\)
=> \(\frac{1}{2017^{2017}}+\frac{1}{2017^{2016}}+\frac{1}{2017^{2015}}+...+\frac{1}{2017}< \frac{1}{2016^{2017}}+\frac{1}{2016^{2016}}+\frac{1}{2016^{2015}}+...+\frac{1}{2016}\)
=> A>B ( vì số bị chia và số chia của A và B đều dương, số bị chia của cả 2 đều là 1, cái nào có số chia nhỏ hơn thì lớn hơn)
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