Cho \(M = \sqrt {\dfrac{9}{{16}}}  + 0,\left( 3 \right) + \dfrac{1}{{2020}} + {\left( {\dfrac{1}{2}} \right)^2} + \dfrac{{ - 1}}{3}\) và \(N = \dfrac{{25}}{4} - \dfrac{3}{8}:0,75 + \dfrac{1}{8}.\sqrt {{4^2}}  - \left| { - 5} \right|\). Tính \(M + N\).

Ta có:

\(\begin{array}{l}M = \sqrt {\dfrac{9}{{16}}}  + 0,\left( 3 \right) + \dfrac{1}{{2020}} + {\left( {\dfrac{1}{2}} \right)^2} + \dfrac{{ - 1}}{3}\\M = \dfrac{3}{4} + \dfrac{3}{9} + \dfrac{1}{{2020}} + \dfrac{1}{4} + \dfrac{{ - 1}}{3}\\M = \left( {\dfrac{3}{4} + \dfrac{1}{4}} \right) + \left( {\dfrac{1}{3} + \dfrac{{ - 1}}{3}} \right) + \dfrac{1}{{2020}}\\M = 1 + 0 + \dfrac{1}{{2020}}\\M = \dfrac{{2020}}{{2020}} + \dfrac{1}{{2020}} = \dfrac{{2021}}{{2020}}\end{array}\)

\(\begin{array}{l}N = \dfrac{{25}}{4} - \dfrac{3}{8}:0,75 + \dfrac{1}{8}.\sqrt {{4^2}}  - \left| { - 5} \right|\\N = \dfrac{{25}}{4} - \dfrac{3}{8}:\dfrac{3}{4} + \dfrac{1}{8}.4 - 5\\N = \dfrac{{25}}{4} - \dfrac{1}{2} + \dfrac{1}{2} - 5\\N = \left( { - \dfrac{1}{2} + \dfrac{1}{2}} \right) + \dfrac{{25}}{4} - 5\\N = 0 + \dfrac{{25}}{4} - 5\\N = \dfrac{{25}}{4} - \dfrac{{20}}{4} = \dfrac{5}{4}\end{array}\)

\(M + N = \dfrac{{2021}}{{2020}} + \dfrac{5}{4} = \dfrac{{2021}}{{2020}} + \dfrac{{2525}}{{2020}} = \dfrac{{4546}}{{2020}} = \dfrac{{2273}}{{1010}}\).