Solve frac{3sqrt{2}}{sqrt{8}}-frac{sqrt{3}}{sqrt{2}-1}+2/3-sqrt{3}

$$\frac { 3 \sqrt { 2 } } { \sqrt { 8 } } - \frac { \sqrt { 3 } } { \sqrt { 2 } - 1 } + \frac { 2 } { 3 - \sqrt { 3 } }$$

Evaluate

$$-\frac{2\sqrt{3}}{3}+\frac{5}{2}-\sqrt{6}\approx -1.104190281$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Cancel out \sqrt{2} in both numerator and denominator.

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

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

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

Consider \left(\sqrt{2}-1\right)\left(\sqrt{2}+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}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

Square 3. Square \sqrt{3}.

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

Subtract 3 from 9 to get 6.

$$\frac{3}{2}-\sqrt{3}\left(\sqrt{2}+1\right)+\frac{2\left(3+\sqrt{3}\right)}{6}$$

Divide 2\left(3+\sqrt{3}\right) by 6 to get \frac{1}{3}\left(3+\sqrt{3}\right).

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

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

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

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

$$\frac{3}{2}-\left(\sqrt{6}+\sqrt{3}\right)+\frac{1}{3}\left(3+\sqrt{3}\right)$$

To find the opposite of \sqrt{6}+\sqrt{3}, find the opposite of each term.

$$\frac{3}{2}-\sqrt{6}-\sqrt{3}+\frac{1}{3}\left(3+\sqrt{3}\right)$$

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

$$\frac{3}{2}-\sqrt{6}-\sqrt{3}+\frac{1}{3}\times 3+\frac{1}{3}\sqrt{3}$$

Cancel out 3 and 3.

$$\frac{3}{2}-\sqrt{6}-\sqrt{3}+1+\frac{1}{3}\sqrt{3}$$

Convert 1 to fraction \frac{2}{2}.

$$\frac{3}{2}-\sqrt{6}-\sqrt{3}+\frac{2}{2}+\frac{1}{3}\sqrt{3}$$

Since \frac{3}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

$$\frac{3+2}{2}-\sqrt{6}-\sqrt{3}+\frac{1}{3}\sqrt{3}$$

Add 3 and 2 to get 5.

$$\frac{5}{2}-\sqrt{6}-\sqrt{3}+\frac{1}{3}\sqrt{3}$$

Combine -\sqrt{3} and \frac{1}{3}\sqrt{3} to get -\frac{2}{3}\sqrt{3}.

$$\frac{5}{2}-\sqrt{6}-\frac{2}{3}\sqrt{3}$$