I'm solving quite an old book of mathematical problems. This one:
Yegerev V.K., Zaytsev V.V., Kordemsky B.A. and others. Sbornik zadach po matematike dlya postupayuschih vo vtuzi. Kniga 1. Algebra (Math problem book for enrollees in technical colleges. Book 1. Algebra) Edited by Skanavi M.I. - 6th edition.
It's famous among Russian-speaking people for giving a good math challenge to wits, however, it sometimes has typos. I've stumbled upon a task #1.224 with a typo which I am unable to figure out. The task is to simplify the following expression:
$$\left(\dfrac{2(a+1)+2\sqrt{a^2+2a}}{3a+1-2\sqrt{2a^2+a}}\right)^{\frac12}-\left(\sqrt{2a+1}-\sqrt{a}\right)^{-1}\sqrt{a+2}$$
The answer in the book is $\dfrac{\sqrt{a}}{\left(\sqrt{2a+1}-\sqrt{a}\right)}$, however, neither I myself can't get the answer from the problem, nor Wolfram Alpha was able to.
Please help me to sort this out, finding and eliminating the typo. I would also like to know how to 'brute force' the expression to find and eliminate such typos.
Thanks.
Could be that Wolfram Alpha is trying too hard to account for a complex $a$. In real numbers, obviously assuming $a\ge 0$, your expression is this:
$$\begin{array}{rcl}\left(\frac{2(a+1)+2\sqrt{a^2+2a}}{3a+1-2\sqrt{2a^2+a}}\right)^\frac{1}{2}-\frac{\sqrt{a+2}}{\sqrt{2a+1}-\sqrt{a}}&=&\left(\frac{(a+2)+a+2\sqrt{a(a+2)}}{(2a+1)+a-2\sqrt{a(2a+1)}}\right)^\frac{1}{2}-\frac{\sqrt{a+2}}{\sqrt{2a+1}-\sqrt{a}}\\&=&\left(\frac{(\sqrt{a+2}+\sqrt{a})^2}{(\sqrt{2a+1}-\sqrt{a})^2}\right)^\frac{1}{2}-\frac{\sqrt{a+2}}{\sqrt{2a+1}-\sqrt{a}}\\&=&\left(\left(\frac{\sqrt{a+2}+\sqrt{a}}{\sqrt{2a+1}-\sqrt{a}}\right)^2\right)^\frac{1}{2}-\frac{\sqrt{a+2}}{\sqrt{2a+1}-\sqrt{a}}\\&=&\frac{\sqrt{a+2}+\sqrt{a}}{\sqrt{2a+1}-\sqrt{a}}-\frac{\sqrt{a+2}}{\sqrt{2a+1}-\sqrt{a}}\\&=&\frac{\sqrt{a+2}+\sqrt{a}-\sqrt{a+2}}{\sqrt{2a+1}-\sqrt{a}}\\&=&\frac{\sqrt{a}}{\sqrt{2a+1}-\sqrt{a}}\end{array}$$
Note that $a\ge 0$ also means $2a+1\gt a$, so the expression $\frac{\sqrt{a+2}+\sqrt{a}}{\sqrt{2a+1}-\sqrt{a}}$ is positive and so the square and the square root can cancel each other. (In general, $(x^2)^\frac{1}{2}=|x|$, which is the same as $x$ if we know that it is positive.)