Rationalize and simplify $\frac{x-1}{\sqrt{2\sqrt{x}} + 1 - \sqrt[4]{x}}$ For my exercise, I have been asked to rationalize and simplify this surd;
$$\frac{x-1}{\sqrt{2\sqrt{x}} + 1 - \sqrt[4]{x}}$$
I don't know how to type it. The denominator is square root of 2 with square root of x after that is + 1 - fourth root of x.]
Each time I do this I get the wrong answer. The method I am using is;
  1133√−7×33√−733√−7

I'm confused with the $\sqrt{2}$,where $\sqrt{x}$ is inside the $\sqrt{2}$ and $1$ separated from $\sqrt{2}$ and $\sqrt{x}$. Sorry if I can't really elaborate it correctly.
This ends up nowhere near the right answer, even once it is simplified, can someone tell me where i'm going wrong?
Many thanks!
 A: Note that $\sqrt{2\sqrt x}=\sqrt2\root4\of x$, so your denominator (if I have understood your comments) is $1+(\sqrt2-1)\root4\of x$. Can you rationalize now? 
A: It seems to me that there are (at least) two possible problems we can consider. The first problem is to rewrite the expression with a polynomial denominator, and the second problem is to rewrite the expression with an integer-coefficient polynomial denominator. I'll first take care of the first problem and then continue so as to take care of the second problem.
$$\frac{x-1}{\sqrt{2\sqrt x\;}+1-\sqrt[4] x}\;\;=\;\;\frac{x-1}{1+(\sqrt 2-1)\sqrt[4] x}\;\;=\;\;\frac{x-1}{1-(1-\sqrt 2)\sqrt[4] x}$$
Using the identity $1-u^4=(1-u)\left(1+u+u^2+u^3\right)$ with $u=(1-\sqrt 2)\sqrt[4] x,$ we can multiply both the numerator and the denominator of the right-most displayed fraction above to get
$$\frac{(x-1)\left(1+u+u^2+u^3\right)}{1-(1-\sqrt 2)^4x},$$
which takes care of the first problem. To solve the second problem, first note that the last fraction is equal to
$$\frac{(x-1)\left(1+u+u^2+u^3\right)}{(1+17x)-12x\sqrt{2}}$$
Now simply multiply both the numerator and the denominator by $(1+17x)+12x\sqrt{2},$ and we're done.
A: As Gerry wrote in his answer:
$$\frac{x-1}{\sqrt{2\sqrt x}+1-\sqrt[4] x}=\frac{x-1}{1+(\sqrt 2-1)\sqrt[4] x}=\frac{x}{1+(\sqrt 2-1)\sqrt[4] x}-\frac{1}{1+(\sqrt 2-1)\sqrt[4] x}$$
Can you take it from here?
A: Hint $ $ Let $\rm\: z\, =\, (1\!-\!\sqrt 2)\sqrt[4]x,\ $ so $\rm\:z^2 = (3\!-\!2\sqrt2)\sqrt x,\,$ $\rm\:z^4\, =\, (17\!-\!12\sqrt2)\,x\, =:\, \alpha\,x,\ $ hence
$$\begin{eqnarray}\rm \frac{1}{1-(1\!-\!\sqrt 2)\sqrt[4]x} &\ =\ &\rm  \frac{1}{1-z}\ =\ \frac{1+z\,\ }{1-z^2}\ =\ \frac{(1+z)\,(1+z^2)}{1- z^4}\\ &\ =\ &\rm \frac{(1+z)\,(1+z^2)}{1-\alpha \,x}\ =\ \frac{(1+z)\,(1+z^2)(1-\alpha'x)}{1-(\alpha\!+\!\alpha')\,x+\alpha\alpha'\,x^2}\end{eqnarray} $$
In our case we have $\rm\, \alpha\!+\!\alpha' = 34,\ \alpha\alpha' = 1,\,$ so the denominator is $\rm\:1 - 34\,x + x^2.$ The final result is obtained by multiplying by $\rm\:x\!-\!1\:$ and substituting in the above the actual values of $\rm\:z\:$ and $\rm\:z^2.$
