Answer
False.
Work Step by Step
The given expression is
$\frac{\sqrt3+7}{\sqrt3-2}=-\frac{7}{2}$
Solve left hand side.
$=\frac{\sqrt3+7}{\sqrt3-2}$
The conjugate of the denominator is $\sqrt3+2$.
Multiply the numerator and the denominator by $\sqrt3+2$.
$=\frac{\sqrt3+7}{\sqrt3-2}\cdot \frac{\sqrt3+2}{\sqrt3+2}$
Use FOIL method in the numerator and the special formula $(A-B)(A+B)=A^2-B^2$ in the denominator.
$=\frac{\sqrt3\cdot \sqrt3+2\cdot \sqrt3+7\cdot \sqrt3+7\cdot 2}{(\sqrt3)^2-(2)^2}$
Use product rule.
$=\frac{3+2\sqrt3+7 \sqrt3+14}{3-4}$
Add like terms.
$=\frac{17+9\sqrt3}{-1}$
$=-(17+9\sqrt3)$.
Hence, the statement is false.
