Trả lời câu hỏi 4 Bài 6 trang 21 SGK Toán 7 Tập 1

Tính:\(\dfrac{{{{72}^2}}}{{{{24}^2}}};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{{{\left( { - 7,5} \right)}^3}}}{{{{\left( {2,5} \right)}^3}}};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{{{15}^3}}}{{27}}\)

Lời giải

\(\eqalign{
& {{{{72}^2}} \over {{{24}^2}}} = {\left( {{{72} \over {24}}} \right)^2} = {\left( 3 \right)^2} = 9\, \cr
& \,{{{{\left( { - 7,5} \right)}^3}} \over {{{\left( {2,5} \right)}^3}}} = {\left( {{{ - 7,5} \over {2,5}}} \right)^3} = {\left( { - 3} \right)^3} = - 27 \cr
& {{{{15}^3}} \over {27}} = {{{{15}^3}} \over {{3^3}}} = {\left( {{{15} \over 3}} \right)^3} = {\left( 5 \right)^3} = 125 \cr} \)


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