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Bài toán : So sánh A và B
\(A=\frac{2018^{100}}{1+2018+2018^2+...+2018^{100}}\)
+) Ta có \(\frac{1}{A}=\frac{1+2018+2018^2+...+2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{2018}{2018^{100}}+\frac{2018^2}{2018^{100}}+...+\frac{2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1\)
\(B=\frac{2019^{100}}{1+2019+2019^2+...+2019^{100}}\)
+) Ta có \(\frac{1}{B}=\frac{1+2019+2019^2+...+2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{2019}{2019^{100}}+\frac{2019^2}{2019^{100}}+...+\frac{2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
+) \(\frac{1}{2018^{100}}>\frac{1}{2019^{100}}\)
\(\frac{1}{2018^{99}}>\frac{1}{2019^{99}}\)
.....................................
\(1=1\)
\(\Rightarrow\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1>\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
\(\Rightarrow\frac{1}{A}>\frac{1}{B}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
\(A=\frac{2018^{2019}-1}{2018^{2019}+1}=\frac{2018^{2019}+1-2}{2018^{2019}+1}=\frac{2018^{2019}+1}{2018^{2019}+1}-\frac{2}{2018^{2019}+1}=1-\frac{2}{2018^{2019}+1}\)
\(B=\frac{2018^{2019}}{2018^{2019}+2}=\frac{2018^{2019}+2-2}{2018^{2019}+2}=\frac{2018^{2019}+2}{2018^{2019}+2}-\frac{2}{2018^{2019}+2}=1-\frac{2}{2018^{2019}+2}\)
Ta có: \(\frac{2}{2018^{2019}+1}>\frac{2}{2018^{2019}+2}\)
\(\Rightarrow1-\frac{2}{2018^{2019}+1}< 1-\frac{2}{2018^{2019}+2}\)
\(\Rightarrow A< B\)
Vậy .....
Ta có : \(0< \frac{2017}{2018}< 1\) nên \(\frac{2017}{2018}>\frac{2017+2019}{2018+2019}\)(1)
\(0< \frac{2018}{2019}< 1\) nên \(\frac{2018}{2019}>\frac{2018+2018}{2018+2019}\) (2)
Cộng vế theo vế 1 và 2 ta được : \(B=\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017+2018+2018+2019}{2018+2019}=\frac{2017+2018}{2018 +2019}+1=A+1>A\)
Vậy B>A
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: \(\frac{2018}{2018^{2019}-2017}>\frac{2018}{2018^{2019}-2016}\)
\(\Rightarrow1+\frac{2018}{2018^{2019}-2017}>1+\frac{2018}{2018^{2019}-2016}\\ \Rightarrow A>B\)
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\)
Với : \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\)
Và : \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\)
\(B=1-\frac{1}{2020}< 1< A\)
