Question

Represent the following mixed infinite decimal periodic fractions as common fractions:
Simplify the following expressions.
$(\frac{a^2-8}{a^2-5a+6}-\frac{(a+1{)}^2+3}{a-3}+\frac{a^2+a}{\sqrt[4]{4a}}):\frac{\sqrt{ab}}{\sqrt[4]{4a^2{b}^2}}$

Answer 1

Given expressions

$=(\frac{a^2-8}{a^2-5a+6}-\frac{(a+1{)}^2+3}{a-3}+\frac{a^2+a}{\sqrt[4]{4a}}):\frac{\sqrt{ab}}{\sqrt[4]{4a^2{b}^2}}$

$=(\frac{a^2-8}{(a-2)(a-3)}-\frac{(a^2-2a+4)}{a-3}+\frac{a(a+1)}{\sqrt[4]{4a}}):\frac{\sqrt{ab}}{\sqrt{ab}}$

$=\frac{\sqrt[4]{4a}(a^2-8)\sqrt[4]{4a}(a-2)(a^2+2a+4)+a(a+1)(a-2)(a-3)}{(a-2)(a-3)\left(\sqrt[4]{4a}\right)}$

$=\frac{\sqrt[4]{4a}[a^2-8(a-2\left)\right(a^2+2a+4\left)\right]+a(a+1)(a-2)(a-3)}{(a-2)(a-3)\sqrt[4]{4a}}$

$=\frac{\sqrt[a]{aa}[a^2-8-[a^2+2a^2+4a-2a^2-4a-8\left]\right]a(a+1)(a-2)(a-3)}{(a-2)(a-3)\sqrt[4]{4a}}$

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

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

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