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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 ~
Có: \(A=\frac{2018^{2019}+1}{2018^{2019}-2017}=\frac{2018^{2019}+1-2018+2018}{2018^{2019}-2017}=\frac{2018^{2019}-2017+2018}{2018^{2019}-2017}=1+\frac{2018}{2018^{2019}-2017}\)
\(B=\frac{2018^{2019}+2}{2018^{2019}-2016}=\frac{2018^{2019}+2-2018+2018}{2018^{2019}-2016}=\frac{2018^{2019}-2016+2018}{2018^{2019}-2016}=1+\frac{2018}{2018^{2019}-2016}\)
Mà: \(\frac{2018}{2018^{2019}-2017}>\frac{2018}{2018^{2019}-2016}\)
\(\Rightarrow1+\frac{2018}{2018^{2019}-2017}>1+\frac{2018}{2018^{2019}-2016}\\ \Rightarrow A>B\)
Ta có :
\(\frac{2017}{2018}>\frac{2017}{2018+2019}\)
\(\frac{2018}{2019}>\frac{2018}{2018+2019}\)
\(\Rightarrow\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
\(\Rightarrow\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017+2018}{2018+2019}\)
\(\Rightarrow A>B\)
Chúc bạn học tốt !!!!
Vì \(\frac{2017}{2018}>\frac{2017}{2018+2019}\)
Vì \(\frac{2018}{2019}>\frac{2018}{2018+2019}\)
\(\Rightarrow\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017+2018}{2018+2019}\)
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 ~