一道简单初中数学题1

题意

$p=\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}$，求证：$1<p<2$

证明:

$p\approx 1.8$，证毕(大雾

$1<p$挺显然的就不证明了

接下来证明$p<2$

$p<2$等价于$\sqrt[3]{9}-\sqrt[3]{8}<\sqrt[3]{6}-\sqrt[3]{4}$

$9-8=(\sqrt[3]{9}-\sqrt[3]{8})\times ({\sqrt[3]{9}}^2+{\sqrt[3]{8}}^2+\sqrt[3]{72})$

$6-4=(\sqrt[3]{6}-\sqrt[3]{4})\times ({\sqrt[3]{6}}^2+{\sqrt[3]{4}}^2+\sqrt[3]{24})$

$\sqrt[3]{9}-\sqrt[3]{8}=\frac{1}{({\sqrt[3]{9}}^2+{\sqrt[3]{8}}^2+\sqrt[3]{72})}=\frac{1}{(\sqrt[3]{8}1+\sqrt[3]{6}4+\sqrt[3]{72})}$

$\sqrt[3]{6}-\sqrt[3]{4}=\frac{1}{({\sqrt[3]{6}}^2+{\sqrt[3]{4}}^2+\sqrt[3]{24})}=\frac{1}{(\sqrt[3]{3}6+\sqrt[3]{1}6+\sqrt[3]{24})}$

$(\sqrt[3]{8}1+\sqrt[3]{6}4+\sqrt[3]{72})>(\sqrt[3]{3}6+\sqrt[3]{1}6+\sqrt[3]{24})$

$\frac{1}{(\sqrt[3]{8}1+\sqrt[3]{6}4+\sqrt[3]{72})}<\frac{1}{(\sqrt[3]{3}6+\sqrt[3]{1}6+\sqrt[3]{24})}$