Solve 4/1+sqrt{3}-2/sqrt{3-1}+frac{3sqrt{3}}{3+2sqrt{3}}

$$\frac { 4 } { 1 + \sqrt { 3 } } - \frac { 2 } { \sqrt { 3 } - 1 } + \frac { 3 \sqrt { 3 } } { 3 + 2 \sqrt { 3 } }$$

Evaluate

$$3-2\sqrt{3}\approx -0.464101615$$

Solution Steps

Rationalize the denominator of \frac{4}{1+\sqrt{3}} by multiplying numerator and denominator by 1-\sqrt{3}.

$$\frac{4\left(1-\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}-\frac{2}{\sqrt{3}-1}+\frac{3\sqrt{3}}{3+2\sqrt{3}}$$

Consider \left(1+\sqrt{3}\right)\left(1-\sqrt{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}.

$$\frac{4\left(1-\sqrt{3}\right)}{1^{2}-\left(\sqrt{3}\right)^{2}}-\frac{2}{\sqrt{3}-1}+\frac{3\sqrt{3}}{3+2\sqrt{3}}$$

Square 1. Square \sqrt{3}.

$$\frac{4\left(1-\sqrt{3}\right)}{1-3}-\frac{2}{\sqrt{3}-1}+\frac{3\sqrt{3}}{3+2\sqrt{3}}$$

Subtract 3 from 1 to get -2.

$$\frac{4\left(1-\sqrt{3}\right)}{-2}-\frac{2}{\sqrt{3}-1}+\frac{3\sqrt{3}}{3+2\sqrt{3}}$$

Divide 4\left(1-\sqrt{3}\right) by -2 to get -2\left(1-\sqrt{3}\right).

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

Rationalize the denominator of \frac{2}{\sqrt{3}-1} by multiplying numerator and denominator by \sqrt{3}+1.

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

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Cancel out 2 and 2.

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

To find the opposite of \sqrt{3}+1, find the opposite of each term.

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

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

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

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

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

Calculate 3 to the power of 2 and get 9.

$$-2\left(1-\sqrt{3}\right)-\sqrt{3}-1+\frac{3\sqrt{3}\left(3-2\sqrt{3}\right)}{9-\left(2\sqrt{3}\right)^{2}}$$

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

$$-2\left(1-\sqrt{3}\right)-\sqrt{3}-1+\frac{3\sqrt{3}\left(3-2\sqrt{3}\right)}{9-2^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$-2\left(1-\sqrt{3}\right)-\sqrt{3}-1+\frac{3\sqrt{3}\left(3-2\sqrt{3}\right)}{9-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$-2\left(1-\sqrt{3}\right)-\sqrt{3}-1+\frac{3\sqrt{3}\left(3-2\sqrt{3}\right)}{9-4\times 3}$$

Multiply 4 and 3 to get 12.

$$-2\left(1-\sqrt{3}\right)-\sqrt{3}-1+\frac{3\sqrt{3}\left(3-2\sqrt{3}\right)}{9-12}$$

Subtract 12 from 9 to get -3.

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

Cancel out -3 and -3.

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

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

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

Combine 2\sqrt{3} and -\sqrt{3} to get \sqrt{3}.

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

Subtract 1 from -2 to get -3.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

Combine \sqrt{3} and -3\sqrt{3} to get -2\sqrt{3}.

$$-3-2\sqrt{3}+6$$

Add -3 and 6 to get 3.

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