《2025年经纶学典学霸黑白题高中数学必修第一册苏教版》

16. $\boldsymbol{★★★}$ (15分)(2025·江苏无锡高一月考)
(1)计算:$(0.008\ 1)^{-\frac{1}{4}}-\left[3×\left(\dfrac{7}{8}\right)^{0}\right]^{-1}×\left[81^{-0.25}+\left(3\dfrac{3}{8}\right)^{-\frac{1}{3}}\right]^{-\frac{1}{2}}$;
(2)化简:$\dfrac{\left(a^{\frac{2}{3}}· b^{-1}\right)^{-\frac{1}{2}}· a^{-\frac{1}{2}}b^{\frac{1}{3}}}{\sqrt[6]{a· b^{5}}}$.

答案： 16.解：
(1)原式$ = (\frac{3}{10})^{4×(\frac{1}{4})} × (\frac{1}{4})^{-\frac{1}{2}×2} - \frac{1}{3} × [3^{4×(-0.25)} + (\frac{3}{2})^{3×(\frac{1}{3})} × (\frac{1}{3})]^{\frac{1}{2}} = \frac{10}{3} × \frac{1}{3} × (\frac{1}{3} × \frac{2}{3})^{\frac{1}{2}} = \frac{10}{3} × \frac{1}{3} × \frac{\sqrt{2}}{3} = \frac{10\sqrt{2}}{27} = 3$.
(2)原式$ = a^{\frac{1}{3}b^{\frac{1}{2}} · a^{\frac{1}{2}b^{\frac{1}{3}}} ÷ a^{\frac{1}{6}} · b^{\frac{1}{6}} = a^{\frac{1}{3} + \frac{1}{2} - \frac{1}{6}} · b^{\frac{1}{2} + \frac{1}{3} - \frac{1}{6}} = a^{\frac{2}{3}} · b^{\frac{2}{3}} = (ab)^{\frac{2}{3}} = 1$

17. $\boldsymbol{★★★}$ (15分)(2025·江苏泰州高一期中)已知$m=\lg 2$,$10^{n}=3$.
(1)求$10^{\frac{3m-2n}{2}}$的值;
(2)用$m$,$n$表示$\log_{15}20$.

答案： 17.解：
(1)因为$10^{n} = 3$，所以$n = \lg3$，又$m = \lg2$，所以$\frac{3m - 2n}{2} = \frac{1}{2}(3\lg2 - 2\lg3) = \frac{1}{2}(\lg2^{3} - \lg3^{2}) = \frac{1}{2}\lg\frac{8}{9}$，所以$10^{\frac{3m - 2n}{2}} = 10^{\frac{1}{2}\lg\frac{8}{9}} = (10^{\lg\frac{8}{9}})^{\frac{1}{2}} = (\frac{8}{9})^{\frac{1}{2}} = \frac{2\sqrt{2}}{3}$.
(2)因为$n = \lg3,m = \lg2$，所以$\log_{15}20 = \frac{\lg20}{\lg15} = \frac{\lg(2×10)}{\lg(3×5)} = \frac{\lg2 + \lg10}{\lg3 + \lg5} = \frac{\lg2 + 1}{\lg3 + 1 - \lg2} = \frac{m + 1}{n + 1 - m}$.