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