a) \(A=\frac{2+2^2+...+2^{2017}}{1-2^{2017}}\)

Đặt \(B=2+2^2+...+2^{2017}\)

\(\Rightarrow2B=2^2+2^3+...+2^{2018}\)

\(\Rightarrow2B-B=\left(2^2+2^3+...+2^{2018}\right)-\left(2+...+2^{2017}\right)\)

\(\Rightarrow B=2^{2018}-2\)

\(\Rightarrow A=\frac{2^{2018}-2}{1-2^{2017}}\)

\(\Rightarrow A=\frac{-2.\left(1-2^{2017}\right)}{1-2^{2017}}\)

\(\Rightarrow A=-2\)

b)Đề phải là CM: \(A< \frac{2017}{2016^2}\)

 \(A=\frac{1}{2017}+\frac{2}{2017^2}+...+\frac{22017}{2017^{2017}}+\frac{2018}{2017^{2018}}\)

\(\Rightarrow2017A=1+\frac{2}{2017}+...+\frac{22017}{2017^{2016}}+\frac{2018}{2017^{2017}}\)

\(\Rightarrow2017A-A=\left(1+...+\frac{2018}{2017^{2017}}\right)-\left(\frac{1}{2017}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\right)\)

\(\Rightarrow2016A=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}-\frac{2018}{2017^{2018}}\)

Đặt \(\Rightarrow S=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}\)

\(\Rightarrow2017S=2017+1+\frac{1}{2017}+...+\frac{1}{2017^{2016}}\)

\(\Rightarrow2017S-S=\left(2017+1+...+\frac{1}{2017^{2016}}\right)-\left(1+...+\frac{1}{2017^{2017}}\right)\)

\(\Rightarrow2016S=2017-\frac{1}{2017^{2017}}< 2017\)

\(\Rightarrow2016S< 2017\)

\(\Rightarrow S< \frac{2017}{2016}\)

\(\Rightarrow2016A< \frac{2017}{2016}\)

\(\Rightarrow A< \frac{2017}{2016^2}\left(đpcm\right)\)

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)

\(\Leftrightarrow x+2=41\)

\(\Leftrightarrow x=41-2\)

\(\Leftrightarrow x=39\)

???????????????????????????????????????????????????????

Ta có: \(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)

\(=1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)\)

\(=\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2018}\)

\(=2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)

Giờ ta thế vào bài toán ban đầu được

\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)

\(=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}\)

\(=\frac{2017}{2018}\)  

 P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\) 

P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)

P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)

P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)

P\(=\frac{1.51}{50.2}=\frac{51}{100}\)

\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{2017}.\left(1+2+3+...+2017\right)\)

\(S=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{2017}.\frac{\left(1+2017\right).2017}{2}\)

\(S=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2018}{2}\)

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

\(S=\frac{1}{2}.\frac{\left(2+2018\right).2017}{2}\)

\(S=\frac{2020.2017}{4}=505.2017=1018585\)

1. Bài giải:

Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\)

\(\Rightarrow\frac{1}{2}A=A-\frac{1}{2}A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1000}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\right)\)

\(\Rightarrow\frac{1}{2}A=1-\frac{1}{1002}=\frac{1001}{1002}\Rightarrow A=\frac{2002}{1002}=\frac{1001}{501}\)

Vậy \(A=\frac{1001}{501}\)

bài 1.a)\(A=\frac{9^3.25^3}{18^2.125^2}=\frac{3^6.5^6}{2^2.3^4.5^6}=\frac{9}{4}\)

b) \(B=\frac{18}{37}+\frac{19}{37}+\frac{8}{2017}-\frac{4026}{2017}+\frac{2017}{2018}\)

\(=1-\frac{4014}{2017}+\frac{2017}{2018}=\frac{1997}{2017}+\frac{2017}{2018}\)