Question

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

Answer

$$5*sq*r*t-6*sq*r^2*t^2*s*q$$

Solution


Regroup terms.
\[3sqrt-sqrt\times 2sqrt\times 3+sqrt\times 2\]
Take out the constants.
\[3sqrt-(2\times 3)rrttsqsq+sqrt\times 2\]
Simplify  \(2\times 3\)  to  \(6\).
\[3sqrt-6rrttsqsq+sqrt\times 2\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[3sqrt-6{r}^{2}{t}^{2}sqsq+sqrt\times 2\]
Regroup terms.
\[3sqrt-6sq{r}^{2}{t}^{2}sq+sqrt\times 2\]
Regroup terms.
\[3sqrt-6sq{r}^{2}{t}^{2}sq+2sqrt\]
Collect like terms.
\[(3sqrt+2sqrt)-6sq{r}^{2}{t}^{2}sq\]
Simplify.
\[5sqrt-6sq{r}^{2}{t}^{2}sq\]