\(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}\)

\(\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)

\(\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)

\(-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}\)

\(\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}\)

Đặt \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)

\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)

\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)

\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)

\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)

\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)

\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)

\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)

\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)

Đặt \(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{2018^2+2019^2}\)

\(2A=\frac{2}{1^2+2^2}+\frac{2}{2^2+3^2}+\frac{2}{3^2+4^2}+...+\frac{2}{2018^2+2019^2}\)

Có \(a^2+b^2\ge2ab\) ( Cosi cho 2 số dương ) 

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b\)

Mà 1;2;3;4;...;2019 là những số khác nhau nên dấu "=" không xảy ra 

\(\Rightarrow\)\(2A< \frac{2}{2\left(1.2\right)}+\frac{2}{2\left(2.3\right)}+\frac{2}{2\left(3.4\right)}+...+\frac{2}{2\left(2018.2019\right)}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}=1-\frac{1}{2019}< 1\)

\(\Rightarrow\)\(2A< 1\)\(\Rightarrow\)\(A< \frac{1}{2}\) ( đpcm ) 

... 

\(5A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)

\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)

\(\Rightarrow4A=5A-A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Đặt \(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

Khi đó \(4A=B-\frac{99}{5^{100}}< B\)

\(5B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}\)

\(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}+\frac{1}{5^{99}}\)

\(\Rightarrow4B=5B-B=1-\frac{1}{5^{99}}\)

\(\Rightarrow B=\frac{1}{4}-\frac{1}{4\cdot5^{99}}< \frac{1}{4}\)

\(\Rightarrow4A < B\Rightarrow4A< \frac{1}{4}\)

\(\Rightarrow A< \frac{1}{16}\) ( đpcm )

2. \(M=\left(1+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(M=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)

\(\Rightarrow\left(M-N\right)^3=0\)

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~~~

đặt 2²⁰¹⁸ = a ; 3²⁰¹⁹ = b ; 5²⁰²⁰ = c

Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)

\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)

\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)

\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)

\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)

\(B< \frac{3}{4}\)

\(\Rightarrow A>1>\frac{3}{4}>B\)

Mình chỉ biết cách tính B thôi, đây nhé:

B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)

B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)