# Thread: More problems

1. ## More problems

Simplify:

$\displaystyle (2 - a^{2}) \cdot \frac {\sqrt {a + \sqrt {2} }} {\sqrt {(\sqrt {2} + a)^{3}}}, a > \sqrt {2}$

This one, too:

$\displaystyle x^{n-1} \cdot (x^{n})^{2-n}$

2. Originally Posted by p.numminen
Simplify:

$\displaystyle (2 - a^{2}) \cdot \frac {\sqrt {a + \sqrt {2} }} {\sqrt {(\sqrt {2} + a)^{3}}}, a > \sqrt {2}$
$\displaystyle (2 - a^{2}) \frac {\sqrt {a + \sqrt {2} }} {\sqrt {(\sqrt {2} + a)^{3}}}= (2 - a^{2}) \frac {(a + \sqrt {2})^{1/2}} {(\sqrt {2} + a)^{3/2}}= (2 - a^{2}) \frac {1} {(\sqrt {2} + a)}=\sqrt{2}-a$

3. How exactly do you get

$\displaystyle \sqrt{2}-a$ out of $\displaystyle (2 - a^{2}) \frac {1} {(\sqrt {2} + a)}$ ?

4. Hello, p.numminen!

Simplify: .$\displaystyle (2 - a^{2}) \cdot \frac {\sqrt {a + \sqrt {2} }} {\sqrt {(\sqrt {2} + a)^{3}}},\;\;a > \sqrt {2}$

We have: .$\displaystyle (2-a^2)\cdot \frac{(\sqrt{2}+a)^{\frac{1}{2}}} {(\sqrt{2}+a)^{\frac{3}{2}}}\;\;=\;\;\frac{2-a^2}{\sqrt{2}+a}$

The numerator factors: .$\displaystyle \frac{(\sqrt{2} - a)(\sqrt{2} + a)}{\sqrt{2} + a}$. . . . . and we can cancel.