Tính bằng cách hợp lí a) (1/3+12/67+13/41)-(79/67-28/41) b)18/13-0,(3)-(13/9+18/13+1/3) c )2×6^9-2^5×18^4/2^2×6^8 d)15^3+5×15^2-5^3/18^3+6×18^2-6^3

Đáp án:

\(\begin{array}{l} a)\,\,\frac{1}{3}\\ b)\,\, - \frac{{19}}{9}\\ c)\,\,5\\ d)\,\,{\left( {\frac{5}{6}} \right)^3} \end{array}\)

Giải thích các bước giải: \[\begin{array}{l} a)\,\,\left( {\frac{1}{3} + \frac{{12}}{{67}} + \frac{{13}}{{41}}} \right) - \left( {\frac{{79}}{{67}} - \frac{{28}}{{41}}} \right)\\ = \frac{1}{3} - \left( {\frac{{79}}{{67}} - \frac{{12}}{{67}}} \right) + \left( {\frac{{13}}{{41}} + \frac{{28}}{{41}}} \right)\\ = \frac{1}{3} - \frac{{67}}{{67}} + \frac{{41}}{{41}}\\ = \frac{1}{3} - 1 + 1\\ = \frac{1}{3}\\ b)\,\,\frac{{18}}{{13}} - 0,\left( 3 \right) - \left( {\frac{{13}}{9} + \frac{{18}}{{13}} + \frac{1}{3}} \right)\\ = \frac{{18}}{{13}} - \frac{1}{3} - \frac{{13}}{9} - \frac{{18}}{{13}} - \frac{1}{3}\\ = \left( {\frac{{18}}{{13}} - \frac{{18}}{{13}}} \right) - \left( {\frac{1}{3} + \frac{1}{3}} \right) - \frac{{13}}{9}\\ = - \frac{2}{3} - \frac{{13}}{9}\\ = - \frac{6}{9} - \frac{{13}}{9}\\ = - \frac{{19}}{9}\\ c)\,\,\frac{{{{2.6}^9} - {2^5}{{.18}^4}}}{{{2^2}{{.6}^8}}}\\ = \frac{{{{2.6}^4}{{.6}^5} - {2^5}{{.6}^4}{{.3}^4}}}{{{2^2}{{.6}^8}}}\\ = \frac{{{{2.6}^4}\left( {{6^5} - {2^4}{{.3}^4}} \right)}}{{{2^2}{{.6}^8}}}\\ = \frac{{{6^5} - {6^4}}}{{{6^4}}}\\ = \frac{{{6^4}\left( {6 - 1} \right)}}{{{6^4}}} = 5\\ d)\,\,\frac{{{{15}^3} + {{5.15}^2} - {5^3}}}{{{{18}^3} + {{6.18}^2} - {6^3}}}\\ = \frac{{{5^3}{{.3}^3} + {5^3}{{.3}^2} - {5^3}}}{{{6^3}{{.3}^3} + {6^3}{{.3}^2} - {6^3}}}\\ = \frac{{{5^3}\left( {{3^3} + {3^2} - 1} \right)}}{{{6^3}\left( {{3^3} + {3^2} - 1} \right)}}\\ = \frac{{{5^3}}}{{{6^3}}} = {\left( {\frac{5}{6}} \right)^3} \end{array}\]