Rationalize $\left(\sqrt{3x+5}-\sqrt{5x+11} -\sqrt{x+9}\right)^{-1}$ I was trying to find if there a method similar to multiplying and dividing by the conjugate $$\frac{1}{\sqrt{3x+5}-\sqrt{5x+11} - \sqrt{x+9}},$$ but that doesn't seem to work here. Also, is there a method of multiplying by a conjugate for roots other than the square root? Such as $(ax+b)^\frac1n ± (cx+d)^\frac1m$
 A: You can apply it twice:
$$\frac1{\sqrt a-\sqrt b-\sqrt c}=\frac{(\sqrt a-\sqrt b)+\sqrt c}{(\sqrt a-\sqrt b)^2-c}=\frac{\sqrt a-\sqrt b+\sqrt c}{a+b-c-2\sqrt{ab}},$$
then
$$\frac{\sqrt a-\sqrt b+\sqrt c}{a+b-c-2\sqrt{ab}}=\frac{(\sqrt a-\sqrt b+\sqrt c)(a+b-c+2\sqrt{ab})}{(a+b-c)^2-4ab}.$$
A: In case you want to approach the general one, if you see the following in denominator,
$$(p^{1/n}-q^{1/m})$$
First focus on p
$$(p^{1/n}-q^{1/m})=(p^{1/n}-(q^{n/m})^{1/n})$$
multiply that by $p^{(n-1)/n}+p^{(n-2)/n}q^{n/m}+...+(q^{n/m})^{(n-1)/n}$
You get $$p-q^{n/m}=(p^m)^{1/m}-(q^n)^{1/m}$$
then again multiply this by $(p^m)^{(m-1)/m}+...(q^n)^{(m-1)/m}$
you'll get $p^m-q^n$
A: Hint: 
$$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right) \left(\sqrt{a}-\sqrt{b}-\sqrt{c}\right) \left(-\sqrt{a}+\sqrt{b}-\sqrt{c}\right) \left(-\sqrt{a}-\sqrt{b}+\sqrt{c}\right) \\ = a^2+b^2+c^2-2(ab+bc+ca)$$
A: Try this for size:
${{\left(\sqrt{3x+5}-\sqrt{5x+11}+\sqrt{x+9}\right)\left(7x+7+2\sqrt{(3x+5)(5x+11)}\right)}\over{(7x+7)^2-4(3x+5)(5x+11)}}.$    
For the second part of your question, see math.stackexchange.com/questions/838287 .
