Giải:

a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\) 

=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\) 

=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\) 

=\(\dfrac{7}{5}+\dfrac{-1}{4}\) 

=\(\dfrac{23}{20}\) 

b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\) 

=\(\left[0+0\right]:\dfrac{2017}{2018}\) 

=0\(:\dfrac{2017}{2018}\) 

=0

c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)

=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) 

=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10

- Ta có: \(A=\frac{\sqrt{x+1}}{\sqrt{x-1}}\)

- Thay  \(x=\frac{16}{9}\)vào đa thức \(A,\)ta có:

             \(A=\frac{\sqrt{\frac{16}{9}+1}}{\sqrt{\frac{16}{9}-1}}\)

      \(\Leftrightarrow A=\frac{\sqrt{\frac{25}{9}}}{\sqrt{\frac{7}{9}}}\)

      \(\Leftrightarrow A=\frac{5\sqrt{7}}{7}\)

Vậy \(A=\frac{5\sqrt{7}}{7}\)

Thay x = 16/9 vào biểu thức, ta có: 

\(\frac{\sqrt{\frac{16}{9}+1}}{\sqrt{\frac{16}{9}-1}}=\frac{\sqrt{\frac{25}{9}}}{\sqrt{\frac{7}{9}}}=\frac{\frac{5}{3}}{\frac{\sqrt{7}}{3}}=\frac{5\sqrt{7}}{5}\)

Ta có : \(\frac{2-x}{2018}-1=\frac{1-x}{2019}-\frac{x}{2020}\)

=> \(\frac{2-x}{2018}+1=\frac{1-x}{2019}+1-\frac{x}{2020}+1\)

=> \(\frac{2020-x}{2018}=\frac{2020-x}{2019}-\frac{2020-x}{2020}\)

=> \(\left(2020-x\right)\left(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2020}\right)=0\)

=> \(2020-x=0\)

=> \(x=2020\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{2020\right\}\)