Question

Represent the following mixed infinite decimal periodic fractions as common fractions:
Simplify the following expressions.
$(1-a^2):\left(\right(\frac{1-a^{3/2}}{1-a^{1/2}}+a^{1/2}\left)\right(\frac{1+a^{3/2}}{1+a^{1/2}}-a^{1/2}\left)\right)+1.$

Answer 1

$(1-a^2):\left[(\frac{1-a^{3/2}}{1-a^{1/2}}+a^{1/2})(\frac{1+a^{1/2}}{1+a^{1/2}}-a^{1/2})\right]+1$

$=(1-a{)}^2:\left[\left(\frac{1-a^{3/2}+a^{1/2}-a}{1-a^{1/2}}\right)\left(\frac{1+a^{3/2}-a^{1/2}-a}{1+a^{1/2}}\right)\right]+1$

$=(1-a{)}^2:\left[\left(\frac{1-a{\sqrt{a}}^3+\sqrt{a}-a}{(1-\sqrt{a})}\right)\left(\frac{1+{\sqrt{a}}^3-\sqrt{a}-a}{(1+\sqrt{a})}\right)\right]+1$

$=(1-a{)}^2:\left[\left(\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}\right)\left(\frac{1+a\sqrt{a}-\sqrt{a}-a}{1+\sqrt{a}}\right)\right]+1$

$=(1-a{)}^2:\left\{\left(\frac{(1-a)-\sqrt{a}(a-1)}{(1-\sqrt{a})}\right)\left(\frac{(1-\sqrt{a}+a\sqrt{-1})}{(1+\sqrt{a})}\right)\right\}+1$

$(1-a^2):\left\{\left(\frac{(1-a)(1+\sqrt{a})}{(1-\sqrt{a})}\right)\left(\frac{(1-\sqrt{a})(1-a)}{(1+\sqrt{a})}\right)\right\}+1$

$(1-a^2):\left\{\left[\frac{(1-\sqrt{a})(1+\sqrt{a})(1+\sqrt{a})}{(1-\sqrt{a})}\right]\left[\frac{(1-\sqrt{a})(1+\sqrt{a})(1-\sqrt{a})}{(1+\sqrt{a})}\right]\right\}+1$

$=(1-a{)}^2:(1+\sqrt{a}{)}^2.(1-\sqrt{a}{)}^2+1$

$=(1-a^2):(1-a^2)+1$

$=\frac{(1-a)(1+a)}{(1-a)(1-a)}+1$

$=\frac{1+a+1-a}{1-a}$

$=\frac{2}{1-a}$