Solve frac{4sqrt{3}}{2-sqrt{2}}-30/4sqrt{3-sqrt{18}}-frac{sqrt{18}}{3-sqrt{12}}

$$\frac { 4 \sqrt { 3 } } { 2 - \sqrt { 2 } } - \frac { 30 } { 4 \sqrt { 3 } - \sqrt { 18 } } - \frac { \sqrt { 18 } } { 3 - \sqrt { 12 } }$$

Evaluate

$$4\sqrt{6}\approx 9.797958971$$

Short Solution Steps

Rationalize the denominator of \frac{4\sqrt{3}}{2-\sqrt{2}} by multiplying numerator and denominator by 2+\sqrt{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

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

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2^{2}-\left(\sqrt{2}\right)^{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Square 2. Square \sqrt{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{4-2}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Subtract 2 from 4 to get 2.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30}{4\sqrt{3}-3\sqrt{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Rationalize the denominator of \frac{30}{4\sqrt{3}-3\sqrt{2}} by multiplying numerator and denominator by 4\sqrt{3}+3\sqrt{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{\left(4\sqrt{3}-3\sqrt{2}\right)\left(4\sqrt{3}+3\sqrt{2}\right)}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Consider \left(4\sqrt{3}-3\sqrt{2}\right)\left(4\sqrt{3}+3\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{\left(4\sqrt{3}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{4^{2}\left(\sqrt{3}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{16\left(\sqrt{3}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

The square of \sqrt{3} is 3.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{16\times 3-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Multiply 16 and 3 to get 48.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-\left(-3\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

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

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-\left(-3\right)^{2}\left(\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Calculate -3 to the power of 2 and get 9.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-9\left(\sqrt{2}\right)^{2}}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

The square of \sqrt{2} is 2.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-9\times 2}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Multiply 9 and 2 to get 18.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{48-18}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Subtract 18 from 48 to get 30.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\frac{30\left(4\sqrt{3}+3\sqrt{2}\right)}{30}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Cancel out 30 and 30.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-\left(4\sqrt{3}+3\sqrt{2}\right)-\frac{\sqrt{18}}{3-\sqrt{12}}$$

To find the opposite of 4\sqrt{3}+3\sqrt{2}, find the opposite of each term.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-4\sqrt{3}-3\sqrt{2}-\frac{\sqrt{18}}{3-\sqrt{12}}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-4\sqrt{3}-3\sqrt{2}-\frac{3\sqrt{2}}{3-\sqrt{12}}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

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

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

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}.

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

Calculate 3 to the power of 2 and get 9.

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

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

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

Calculate -2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2}-4\sqrt{3}-3\sqrt{2}-\frac{3\sqrt{2}\left(3+2\sqrt{3}\right)}{9-12}$$

Subtract 12 from 9 to get -3.

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

Cancel out -3 and -3.

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

The opposite of -\sqrt{2}\left(3+2\sqrt{3}\right) is \sqrt{2}\left(3+2\sqrt{3}\right).

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

To add or subtract expressions, expand them to make their denominators the same. Multiply -4\sqrt{3}-3\sqrt{2} times \frac{2}{2}.

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

Since \frac{4\sqrt{3}\left(2+\sqrt{2}\right)}{2} and \frac{2\left(-4\sqrt{3}-3\sqrt{2}\right)}{2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 4\sqrt{3}\left(2+\sqrt{2}\right)+2\left(-4\sqrt{3}-3\sqrt{2}\right).

$$\frac{8\sqrt{3}+4\sqrt{6}-8\sqrt{3}-6\sqrt{2}}{2}+\sqrt{2}\left(3+2\sqrt{3}\right)$$

Do the calculations in 8\sqrt{3}+4\sqrt{6}-8\sqrt{3}-6\sqrt{2}.

$$\frac{4\sqrt{6}-6\sqrt{2}}{2}+\sqrt{2}\left(3+2\sqrt{3}\right)$$

Divide each term of 4\sqrt{6}-6\sqrt{2} by 2 to get 2\sqrt{6}-3\sqrt{2}.

$$2\sqrt{6}-3\sqrt{2}+\sqrt{2}\left(3+2\sqrt{3}\right)$$

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

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

Combine -3\sqrt{2} and 3\sqrt{2} to get 0.

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

Combine 2\sqrt{6} and 2\sqrt{6} to get 4\sqrt{6}.

$$4\sqrt{6}$$