证明恒等式：{[5^(1/4)-3+2√5]／[5^(1/4)+3-2√5]}^(1/3)=[5^(1/4)+1]／[5^(1/4)-1]

myyour

不会输入这些数学符号，只好手写发图片了。

王守恩

\(\displaystyle\big(\frac{\sqrt[4]{a}+1}{\sqrt[4]{a}-1}\big)^3\equiv\frac{\sqrt[4]{a^3}+3\sqrt{a}+3\sqrt[4]{a}+1}{\sqrt[4]{a^3}-3\sqrt{a}+3\sqrt[4]{a}-1}\)

\(\displaystyle\equiv\frac{\sqrt[4]{a^3}+6\sqrt{a}+3\sqrt[4]{a}+2a-2a+1-3\sqrt{a}}{\sqrt[4]{a^3}-6\sqrt{a}+3\sqrt[4]{a}-2a+2a-1+3\sqrt{a}}\)

\(\displaystyle\equiv\frac{(\sqrt[4]{a}+2\sqrt{a})(3+\sqrt{a})-(2a-1+3\sqrt{a})}{(\sqrt[4]{a}-2\sqrt{a})(3+\sqrt{a})-(2a-1+3\sqrt{a})}\)

\(\displaystyle\equiv\frac{\sqrt[4]{a}-\frac{2a-1+3\sqrt{a}}{3+\sqrt{a}}+2\sqrt{a}}{\sqrt[4]{a}+\frac{2a-1+3\sqrt{a}}{3+\sqrt{a}}-2\sqrt{a}}\)

\(\displaystyle 联系本题：a=5,\ \ \frac{2a-1+3\sqrt{a}}{3+\sqrt{a}}=3\)

luyuanhong

楼上 王守恩 的解答很好！已收藏。