Lời giải:
a) Sửa đề thành CM \(\frac{a}{b}< \frac{a+m}{b+m}\) mới đúng nhé.

\(\frac{a}{b}-\frac{a+m}{b+m}=\frac{ab+am-b(a+m)}{b(b+m)}=\frac{m(a-b)}{b(m+b)}\)

Do $\frac{a}{b}< 1; b>0\Rightarrow a< b\Rightarrow a-b< 0$

$m>0; b>0\Rightarrow \frac{m}{b(m+b)}>0$

$\Rightarrow \frac{a}{b}-\frac{a+m}{b+m}=\frac{m(a-b)}{b(m+b)}< 0$

$\Rightarrow \frac{a}{b}< \frac{a+m}{b+m}$ (đpcm)

b)

\(0< B=\frac{2020^{2017+1}}{2020^{2018+1}}=\frac{2020^{2018}}{2020^{2019}}< 1\) do $0< 2020^{2018}< 2020^{2019}$

$1>0$

$\Rightarrow B< \frac{2020^{2018}+1}{2020^{2019}+1}$ (theo phần a)

Hay $B< A$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

M=[ 1+1/2018 +1/2 +1/2017 +1/3 +1/2016 +........+1/1009 +1/1010] .2.3.4...2018

M=[2019/2018 =2019/2.2017 +2019/3.2016 +....+2019/1009.1010]­.2.3.....2018

M.=2019.[1/2018 +1/2.2017 +.....+1/1009.1010]­ .2.3....2018 chia het cho 2019

suy ra M chia het cho2019

vay M chia het cho2019

\(\frac{a^4}{2018}+\frac{b^4}{2019}=\frac{1}{4037}\)

\(\Leftrightarrow\frac{2019a^4+2018b^4}{2018\cdot2019}=\frac{a^2+b^2}{2018+2019}\)

\(\Leftrightarrow\left(2018+2019\right)\left(2019a^4+2018b^4\right)=2018\cdot2019\left(a^2+b^2\right)\)

\(\Leftrightarrow2019^2\cdot a^4+2018^2\cdot b^4+2018\cdot2019\cdot a^4+2018\cdot2019b^4=2018\cdot2019\cdot a^2+2018\cdot2019\cdot b^2\)

\(\Leftrightarrow2019^2\cdot a^4+2018^2\cdot b^4=2018\cdot2019\cdot a^2\left(1-a^2\right)+2018\cdot2019\cdot b^2\left(1-b^2\right)\)

\(\Leftrightarrow\left(2019a^2\right)^2+\left(2018b^2\right)^2=2\cdot2018\cdot2019\cdot a^2\cdot b^2\)

\(\Leftrightarrow\left(2019a^2-2018b^2\right)=0\)

\(\Leftrightarrow2019a^2=2018b^2\Leftrightarrow\frac{a^2}{2018}=\frac{b^2}{2019}=\frac{a^2+b^2}{2018+2019}=\frac{1}{4037}\)

\(\Rightarrow\frac{a^{2018}}{2018^{10009}}=\frac{b^{2018}}{2019^{1009}}=\frac{1}{4037^{1009}}\)

\(\Rightarrow P=\frac{2}{4037^{1009}}\)