Show that $P=\sqrt{a^2-2ab+b^2}+\left(\frac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\frac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)$ is rational

Question

Show that the number $$P=\sqrt{a^2-2ab+b^2}+\left(\dfrac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\dfrac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)$$ is a rational number ($P\in\mathbb{Q})$ if $a\in\mathbb{Q},b\in\mathbb{Q},a>0,b>0$ and $a\ne b$. Find the value of $P$ if $a=1.1$ and $b=1.22$.

My try: $$P=\left|a-b\right|+\dfrac{a-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}:\dfrac{b\sqrt{a}+\sqrt{b}\left(a-\sqrt{ab}\right)}{a-\sqrt{ab}}=\\=|a-b|+\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}.\dfrac{a-\sqrt{ab}}{b\sqrt{a}+a\sqrt{b}-\sqrt{b^2a}}.$$

Answer 1

\begin{align} P&=\sqrt{a^2-2ab+b^2}+\left(\dfrac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\dfrac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)=\\ &=|a-b|+\dfrac{\sqrt{ab}}{\sqrt a-\sqrt b}:\dfrac{b\sqrt a+a\sqrt b-b\sqrt a}{a-\sqrt{ab}}=\\ &=|a-b|+\dfrac{\sqrt{ab}}{\sqrt a-\sqrt b}:\dfrac{a\sqrt b}{\sqrt a\left(\sqrt a-\sqrt b\right)}=\\ &=|a-b|+\dfrac{\sqrt{ab}}{\sqrt a-\sqrt b}\cdot\dfrac{\sqrt a\left(\sqrt a-\sqrt b\right)}{a\sqrt b}=\\ &=|a-b|+\dfrac{a\sqrt b\left(\sqrt a-\sqrt b\right)}{a\sqrt b\left(\sqrt a-\sqrt b\right)}=\\ &=|a-b|+1\;. \end{align}

Since $\;a\;$ and $\;b\;$ are rational numbers, also $\;P\;$ is a rational number.

If $\;a=1.1\;$ and $\;b=1.22\;,\;$ the value of $\;P\;$ is

$\begin{align} P&=|a-b|+1=|1.1-1.22|+1=|-0.12|+1=\\ &=0.12+1=1.12 \end{align}$

Answer 2

Just keep chewing on it until it falls apart

$|a-b|+\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}.\dfrac{a-\sqrt{ab}}{b\sqrt{a}+a\sqrt{b}-\sqrt{b^2a}}=$

$|a-b| +\frac {\sqrt {ab}}{\sqrt a - \sqrt b}\cdot \frac {\sqrt a(\sqrt a -\sqrt b)}{b\sqrt a + a\sqrt b - b\sqrt a}=$

$|a-b| + \frac {\sqrt{ab}}{\sqrt a-\sqrt b}\cdot \frac {\sqrt a(\sqrt a-\sqrt b)}{a\sqrt b}=$

$|a-b| + \frac {\sqrt {ab}\sqrt a}{a\sqrt b}=$

$|a-b| + \frac {\sqrt{a^2b}}{\sqrt{a^2b}}=$

$|a-b| + 1$.