Solve frac{sqrt{3}}{2(sqrt{3}+1)+1}

$$\frac { \sqrt { 3 } } { 2 ( \sqrt { 3 } + 1 ) + 1 }$$

Evaluate

$$2-\sqrt{3}\approx 0.267949192$$

Solution Steps

Use the distributive property to multiply 2 by \sqrt{3}+1.

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

Add 2 and 1 to get 3.

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

Rationalize the denominator of \frac{\sqrt{3}}{2\sqrt{3}+3} by multiplying numerator and denominator by 2\sqrt{3}-3.

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

Consider \left(2\sqrt{3}+3\right)\left(2\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}.

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

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

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\sqrt{3}\left(2\sqrt{3}-3\right)}{4\left(\sqrt{3}\right)^{2}-3^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{3}\left(2\sqrt{3}-3\right)}{4\times 3-3^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{\sqrt{3}\left(2\sqrt{3}-3\right)}{12-3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\sqrt{3}\left(2\sqrt{3}-3\right)}{12-9}$$

Subtract 9 from 12 to get 3.

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

Use the distributive property to multiply \sqrt{3} by 2\sqrt{3}-3.

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

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

$$2-\sqrt{3}$$