Cho \(\begin{array}{l} M = \sqrt {\frac{9}{{16}}} + 0,(3) + \frac{1}{{2020}} + {\left( {\frac{1}{2}} \right)^2} + \frac{{ - 1}}{3}\\ N = \frac{{25}}{4} - \frac{3}{8}:0,75 + \frac{1}{8}.\sqrt {{4^2}} - \left| { - 5} \right| \end{array}\) Tính (M + N ).
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiTa có:
\(\begin{array}{*{20}{l}} {M = \sqrt {\frac{9}{{16}}} + 0,\left( 3 \right) + \frac{1}{{2020}} + {{\left( {\frac{1}{2}} \right)}^2} + \frac{{ - 1}}{3}}\\ {M = \frac{3}{4} + \frac{3}{9} + \frac{1}{{2020}} + \frac{1}{4} + \frac{{ - 1}}{3}}\\ {M = \left( {\frac{3}{4} + \frac{1}{4}} \right) + \left( {\frac{1}{3} + \frac{{ - 1}}{3}} \right) + \frac{1}{{2020}}}\\ {M = 1 + 0 + \frac{1}{{2020}}}\\ {M = \frac{{2020}}{{2020}} + \frac{1}{{2020}} = \frac{{2021}}{{2020}}} \end{array}\)
\(\begin{array}{*{20}{l}} {N = \frac{{25}}{4} - \frac{3}{8}:0,75 + \frac{1}{8}.\sqrt {{4^2}} - \left| { - 5} \right|}\\ {N = \frac{{25}}{4} - \frac{3}{8}:\frac{3}{4} + \frac{1}{8}.4 - 5}\\ {N = \frac{{25}}{4} - \frac{1}{2} + \frac{1}{2} - 5}\\ {N = \left( { - \frac{1}{2} + \frac{1}{2}} \right) + \frac{{25}}{4} - 5}\\ {N = 0 + \frac{{25}}{4} - 5}\\ {N = \frac{{25}}{4} - \frac{{20}}{4} = \frac{5}{4}} \end{array}\)
\(\to M + N = \frac{{2021}}{{2020}} + \frac{5}{4} = \frac{{2021}}{{2020}} + \frac{{2525}}{{2020}} = \frac{{4546}}{{2020}} = \frac{{2273}}{{1010}}\)
