Solve 1/2/sqrt{3/2-frac{sqrt{2}}{2}}-(1-frac{sqrt{3}}{3})-1

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

Evaluate

$$\frac{4\sqrt{3}}{3}+\sqrt{2}-2\approx 1.723614639$$

Short Solution Steps

Since \frac{\sqrt{3}}{2} and \frac{\sqrt{2}}{2} have the same denominator, subtract them by subtracting their numerators.

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

Divide \frac{1}{2} by \frac{\sqrt{3}-\sqrt{2}}{2} by multiplying \frac{1}{2} by the reciprocal of \frac{\sqrt{3}-\sqrt{2}}{2}.

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

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

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}.

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

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

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

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

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

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

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

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{\left(2-2\sqrt{3}+2\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{12-8}-\frac{3-\sqrt{3}}{3}$$

Subtract 8 from 12 to get 4.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 3 is 12. Multiply \frac{\left(2-2\sqrt{3}+2\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{4} times \frac{3}{3}. Multiply \frac{3-\sqrt{3}}{3} times \frac{4}{4}.

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

Since \frac{3\left(2-2\sqrt{3}+2\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{12} and \frac{4\left(3-\sqrt{3}\right)}{12} have the same denominator, subtract them by subtracting their numerators.

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

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

$$\frac{12\sqrt{3}+12\sqrt{2}-36-12\sqrt{6}+12\sqrt{6}+24-12+4\sqrt{3}}{12}$$

Do the calculations in 12\sqrt{3}+12\sqrt{2}-36-12\sqrt{6}+12\sqrt{6}+24-12+4\sqrt{3}.

$$\frac{16\sqrt{3}+12\sqrt{2}-24}{12}$$

Factor

$$\frac{3 \sqrt{2} + 4 \sqrt{3} - 6}{3} = 1.723614639131598$$