Question

Indicate the domain of definition of the function and simplify the given expression.
$y=x\frac{1+2(x+4{)}^{-0.5}}{2-(x+4{)}^{0.5}}+(x+4{)}^{0.5}+4(x+4{)}^{-0.5}$

Answer 1

$y=x.\frac{1+2{(x+4)}^{-0.5}}{2-{(x+4)}^{0.5}}+{(x+4)}^{0.5}+4{(x+4)}^{-0.5}$

$=x.\frac{1+\frac{2}{\sqrt{x+4}}}{2-\sqrt{x+4}}+\sqrt{x+4}+\frac{4}{\sqrt{x+4}}$

$=\frac{x.(\sqrt{x+4}+2)}{2\sqrt{x+4}-(x+4)}+\sqrt{x+4}+\frac{4}{\sqrt{x+4}}$

$=\frac{x.(\sqrt{x+4}+2)}{2\sqrt{x+4}(1-\sqrt{x+4})}+\frac{x+8}{\sqrt{x+4}}$

$=\frac{x.(\sqrt{x+4}+2)+2(x+8)(1-\sqrt{x+4})}{2\sqrt{x+4}(1-\sqrt{x+4})}$

$=\frac{x\left(\sqrt{x+4}\right)+2x+2x+16-2x\sqrt{x+4}-16\sqrt{x+4}}{2\sqrt{x+4}(1-\sqrt{x+4})}$

$=\frac{4x+16-\sqrt{x+4}(x+16)}{2\sqrt{x+4}(1-\sqrt{x+4})}$

For domain

$\sqrt{x+4}>0$

$\Rightarrow x+4>0$

$\Rightarrow x>-4$

and $1-\sqrt{x+4}\ne 0$

$\sqrt{x+4}\ne 1$

$x+4\ne 1$

$x\ne 1-4$

$x\ne -3$

$\therefore x>-4$ and $x\ne -3$

$x\in (-4,\infty )-(-3)$

Hence, the answer is $x\in (-4,\infty )-(-3).$