Question

Represent the following mixed infinite decimal periodic fractions as common fractions:
Simplify the following expressions.
${(\frac{3-\sqrt{a}}{9-a}+\frac{1}{3-\sqrt{a}}-6\frac{a^2+162}{729-a^3})}^{-1}+\frac{a(a+9)}{54}\cdot$

Answer 1

${(\frac{3-\sqrt{a}}{9-a}+\frac{1}{3-\sqrt{a}}-\frac{6(a^2+162)}{729-a^3})}^{-1}+\frac{a(a+9)}{54}⟶\left(1\right)$

$\Rightarrow \frac{3-\sqrt{a}}{9-a}=\frac{3-\sqrt{a}}{(3-\sqrt{a})(3+\sqrt{a})}=\frac{1}{3+\sqrt{a}}$

Now, $\frac{1}{3-\sqrt{a}}+\frac{1}{3-\sqrt{a}}=\frac{3-\sqrt{a}+3+\sqrt{a}}{(3+\sqrt{a})(3-\sqrt{a})}=\frac{6}{9-a}$

$\Rightarrow \frac{6(a^2+162)}{729-a^3}=\frac{6a^2+972}{(9-a)(a^2+9a+81)}$

$\Rightarrow {(\frac{3-\sqrt{a}}{9-a}+\frac{1}{3-\sqrt{a}}-\frac{6(a^2+162)}{729-a^3})}^{-1}={(\frac{6}{9-a}-\frac{({6a}^2+972)}{(9-a)(a^2+9a+81)})}^{-1}$

$\Rightarrow {\left(\frac{{6a}^2+54a+486-6a^2-972}{(9-a)(a^2+9a+81)}\right)}^{-1}={\left(\frac{54a-486}{(9-a)(a^2+9a+81)}\right)}^{-1}$

$\Rightarrow {\left(\frac{54(a-9)}{(9-a)(a^2+9a+81)}\right)}^{-1}$

$\Rightarrow {\left(\frac{-54}{a^2+9a+81}\right)}^{-1}$

$\Rightarrow \frac{-(a^2+9a+81)}{54}=A$

$\Rightarrow A+B=\frac{-(a^2+9a)-81}{54}+\frac{a(a+9)}{54}$

$=\frac{-81}{54}$

$=\frac{-3}{2}.$

Hence, solved.