题目内容
10.[$\frac{1}{4}$(0.027${\;}^{\frac{2}{3}}}$+50×0.0016${\;}^{\frac{3}{4}}}$)]${\;}^{-\frac{1}{2}}}$=$\frac{20}{7}$.(用数字作答)分析 由指数幂的运算可得0.027${\;}^{\frac{2}{3}}}$=$((0.3)^{3})^{\frac{2}{3}}$=0.09,0.0016${\;}^{\frac{3}{4}}}$=0.008,从而求得$\frac{1}{4}$(0.027${\;}^{\frac{2}{3}}}$+50×0.0016${\;}^{\frac{3}{4}}}$)=$\frac{49}{400}$,从而求得.
解答 解:∵0.027${\;}^{\frac{2}{3}}}$=$((0.3)^{3})^{\frac{2}{3}}$=0.09,0.0016${\;}^{\frac{3}{4}}}$=0.008,
故$\frac{1}{4}$(0.027${\;}^{\frac{2}{3}}}$+50×0.0016${\;}^{\frac{3}{4}}}$)=$\frac{1}{4}$(0.09+0.4)=$\frac{49}{400}$,
故[$\frac{1}{4}$(0.027${\;}^{\frac{2}{3}}}$+50×0.0016${\;}^{\frac{3}{4}}}$)]${\;}^{-\frac{1}{2}}}$
=$(\frac{49}{400})^{-\frac{1}{2}}$=$\frac{20}{7}$;
故答案为:$\frac{20}{7}$.
点评 本题考查了指数幂的运算的应用,属于基础题.
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