Solve frac{1+frac{sqrt{3}}{3}}{1-1(frac{sqrt{3}}{3})}

$$\frac{ 1+ \frac{ \sqrt{ 3 } }{ 3 } }{ 1-1( \frac{ \sqrt{ 3 } }{ 3 } ) }$$

Evaluate

$$\sqrt{3}+2\approx 3.732050808$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

$$\frac{\frac{3}{3}+\frac{\sqrt{3}}{3}}{1-1\times \frac{\sqrt{3}}{3}}$$

Since \frac{3}{3} and \frac{\sqrt{3}}{3} have the same denominator, add them by adding their numerators.

$$\frac{\frac{3+\sqrt{3}}{3}}{1-1\times \frac{\sqrt{3}}{3}}$$

Express 1\times \frac{\sqrt{3}}{3} as a single fraction.

$$\frac{\frac{3+\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

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

Since \frac{3}{3} and \frac{\sqrt{3}}{3} have the same denominator, subtract them by subtracting their numerators.

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

Divide \frac{3+\sqrt{3}}{3} by \frac{3-\sqrt{3}}{3} by multiplying \frac{3+\sqrt{3}}{3} by the reciprocal of \frac{3-\sqrt{3}}{3}.

$$\frac{\left(3+\sqrt{3}\right)\times 3}{3\left(3-\sqrt{3}\right)}$$

Cancel out 3 in both numerator and denominator.

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

Rationalize the denominator of \frac{\sqrt{3}+3}{-\sqrt{3}+3} by multiplying numerator and denominator by -\sqrt{3}-3.

$$\frac{\left(\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)}{\left(-\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)}$$

Consider \left(-\sqrt{3}+3\right)\left(-\sqrt{3}-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)}{\left(-\sqrt{3}\right)^{2}-3^{2}}$$

Expand \left(-\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)}{\left(-1\right)^{2}\left(\sqrt{3}\right)^{2}-3^{2}}$$

Calculate -1 to the power of 2 and get 1.

$$\frac{\left(\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)}{1\left(\sqrt{3}\right)^{2}-3^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)}{1\times 3-3^{2}}$$

Multiply 1 and 3 to get 3.

$$\frac{\left(\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)}{3-3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)}{3-9}$$

Subtract 9 from 3 to get -6.

$$\frac{\left(\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)}{-6}$$

Apply the distributive property by multiplying each term of \sqrt{3}+3 by each term of -\sqrt{3}-3.

$$\frac{-\left(\sqrt{3}\right)^{2}-3\sqrt{3}-3\sqrt{3}-9}{-6}$$

The square of \sqrt{3} is 3.

$$\frac{-3-3\sqrt{3}-3\sqrt{3}-9}{-6}$$

Combine -3\sqrt{3} and -3\sqrt{3} to get -6\sqrt{3}.

$$\frac{-3-6\sqrt{3}-9}{-6}$$

Subtract 9 from -3 to get -12.

$$\frac{-12-6\sqrt{3}}{-6}$$

Divide each term of -12-6\sqrt{3} by -6 to get 2+\sqrt{3}.

$$2+\sqrt{3}$$

Factor

$$\sqrt{3} + 2 = 3.732050808$$