Answer
$\frac{2+6\sqrt{3x}+\sqrt{7x}+3x\sqrt{21}}{1-9x}$.
Work Step by Step
The given expression is
$=\frac{4+2\sqrt{7x}}{2-3\sqrt{12x}}$
Multiply and divide by the conjugate of the denominator.
$=\frac{4+2\sqrt{7x}}{2-3\sqrt{12x}}\cdot \frac{2+3\sqrt{12x}}{2+3\sqrt{12x}}$
Use the FOIL method.
$=\frac{8+12\sqrt{12x}+4\sqrt{7x}+(2\sqrt{7x})\cdot (3\sqrt{12x})}{4+6\sqrt{12x}-6\sqrt{12x}+(-3\sqrt{12x})(3\sqrt{12x})}$
Simplify.
$=\frac{8+12\sqrt{4\cdot3x}+4\sqrt{7x}+(2\sqrt{7x})\cdot (3\sqrt{4\cdot3x})}{4-36x}$
$=\frac{8+12\cdot 2\sqrt{3x}+4\sqrt{7x}+(2\sqrt{7x})\cdot (3\cdot 2\sqrt{3x})}{4-36x}$
$=\frac{8+24\sqrt{3x}+4\sqrt{7x}+12x\sqrt{7\cdot 3}}{4-36x}$
Divide the numerator and the denominator by $4$.
$=\frac{2+6\sqrt{3x}+\sqrt{7x}+3x\sqrt{21}}{1-9x}$.
