Solve frac{sqrt{3}+3}{3sqrt{3}-3}

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

Evaluate

$$\frac{2\sqrt{3}}{3}+1\approx 2.154700538$$

Solution Steps

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

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

Consider \left(3\sqrt{3}-3\right)\left(3\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}.

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

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{3} is 3.

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

Multiply 9 and 3 to get 27.

$$\frac{\left(\sqrt{3}+3\right)\left(3\sqrt{3}+3\right)}{27-3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(\sqrt{3}+3\right)\left(3\sqrt{3}+3\right)}{27-9}$$

Subtract 9 from 27 to get 18.

$$\frac{\left(\sqrt{3}+3\right)\left(3\sqrt{3}+3\right)}{18}$$

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

$$\frac{3\left(\sqrt{3}\right)^{2}+3\sqrt{3}+9\sqrt{3}+9}{18}$$

The square of \sqrt{3} is 3.

$$\frac{3\times 3+3\sqrt{3}+9\sqrt{3}+9}{18}$$

Multiply 3 and 3 to get 9.

$$\frac{9+3\sqrt{3}+9\sqrt{3}+9}{18}$$

Combine 3\sqrt{3} and 9\sqrt{3} to get 12\sqrt{3}.

$$\frac{9+12\sqrt{3}+9}{18}$$

Add 9 and 9 to get 18.

$$\frac{18+12\sqrt{3}}{18}$$

Factor

$$\frac{\sqrt{3} {(\sqrt{3} + 2)}}{3} = 2.1547005383792515$$