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ta có A=\(\frac{2017^{2017}+1}{2017^{2018}+1}\)=> 2017A =\(\frac{2017^{2018}+2017}{2017^{2018}+1}=1+\frac{2016}{2017^{2018}+1}\)(1)
B=\(\frac{2017^{2018}+1}{2017^{2019}+1}\)=> 2017B =\(\frac{2017^{2019}+2017}{2017^{2019}+1}=1+\frac{2016}{2017^{2019}+1}\)(2)
So sánh (1)với (2) ta thấy 2017A>2017B
=>A>B
Vậy A>B
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{2017^{2018}+1}{2017^{2019}+1}< \frac{2017^{2018}+1+2016}{2017^{2019}+1+2016}=\frac{2017^{2018}+2017}{2017^{2017}+2017}=\frac{2017\left(2017^{2017}+1\right)}{2017\left(2017^{2016}+1\right)}=A\)
\(\Rightarrow\)\(B< A\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
\(+)A=\frac{10^{2016}+2018}{10^{2017}+2018}\)
\(10A=\frac{10^{2017}+20180}{10^{2017}+2018}=1+\frac{18162}{10^{2017}+2018}\left(1\right)\)
\(+)10B=\frac{10^{2018}+20180}{10^{2018}+2018}=1+\frac{18162}{10^{2018}+2018}\left(2\right)\)
Từ (1),(2)=> \(\frac{18162}{10^{2017}+2018} >\frac{18162}{10^{2018}+2018}\)
=> 10A>10B
=>A>B
\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)
Có \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\)
\(\Rightarrow A< B\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Ta có:
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Cộng vế theo vế, ta có:
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Vậy A > B
Ta có :
\(B=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
Vì :
\(\frac{2017}{2018}>\frac{2017}{2018+2019}\)
\(\frac{2018}{2019}>\frac{2018}{2018+2019}\)
\(\Rightarrow\)\(\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017+2018}{2018+2019}\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
Ta có
A= \(\frac{2017^{2018}-3+4}{2017^{2018}-3}=1+\frac{4}{2017^{2018}-3}\)
B= \(1+\frac{4}{2017^{2018}-5}\)
vậy A > B