Question

rationalize root six divided by root three minus root two. Csqrt%7B3%7D-%5Csqrt%7B2%7D%20%7D" id="TexFormula1" title="\frac{\sqrt{6} }{\sqrt{3}-\sqrt{2} }" alt="\frac{\sqrt{6} }{\sqrt{3}-\sqrt{2} }" align="absmiddle" class="latex-formula">

Answer 1

Answer:

the answer is

$3 \sqrt{2} + 2 \sqrt{3}$

Step-by-step explanation:

the explanation is given in the image.

Answer 2

Answer:

$\huge\boxed{\dfrac{\sqrt6}{\sqrt3-\sqrt2}=3\sqrt2+2\sqrt3}$

Step-by-step explanation:

$\dfrac{\sqrt6}{\sqrt3-\sqrt2}\\\\\text{use}\ (a-b)(a+b)=a^2-b^2\\\\\dfrac{\sqrt6}{\sqrt3-\sqrt2}\cdot\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=\dfrac{\sqrt6(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}=\dfrac{(\sqrt6)(\sqrt3)+(\sqrt6)(\sqrt2)}{(\sqrt3)^2-(\sqrt2)^2}$

$\text{use}\ \sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\ \text{and}\ (\sqrt{a})^2=a$

$=\dfrac{\sqrt{(6)(3)}+\sqrt{(6)(2)}}{3-2}=\dfrac{\sqrt{18}+\sqrt{12}}{1}=\sqrt{9\cdot2}+\sqrt{4\cdot3}\\\\=\sqrt9\cdot\sqrt2+\sqrt4\cdot\sqrt3=3\sqrt2+2\sqrt3$