Solve 3/sqrt{2-1}-frac{sqrt{2}*(6-3sqrt{8})}{2}

$$\frac { 3 } { \sqrt { 2 } - 1 } - \frac { \sqrt { 2 } \cdot ( 6 - 3 \sqrt { 8 } ) } { 2 }$$

Evaluate

$$9$$

Short Solution Steps

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

$$\frac{3\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{\sqrt{2}\left(6-3\sqrt{8}\right)}{2}$$

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}.

$$\frac{3\left(\sqrt{2}+1\right)}{\left(\sqrt{2}\right)^{2}-1^{2}}-\frac{\sqrt{2}\left(6-3\sqrt{8}\right)}{2}$$

Square \sqrt{2}. Square 1.

$$\frac{3\left(\sqrt{2}+1\right)}{2-1}-\frac{\sqrt{2}\left(6-3\sqrt{8}\right)}{2}$$

Subtract 1 from 2 to get 1.

$$\frac{3\left(\sqrt{2}+1\right)}{1}-\frac{\sqrt{2}\left(6-3\sqrt{8}\right)}{2}$$

Anything divided by one gives itself.

$$3\left(\sqrt{2}+1\right)-\frac{\sqrt{2}\left(6-3\sqrt{8}\right)}{2}$$

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}.

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

Multiply -3 and 2 to get -6.

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

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

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

Use the distributive property to multiply \sqrt{2} by 6-6\sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Multiply -6 and 2 to get -12.

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

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

$$\frac{2\left(3\sqrt{2}+3\right)}{2}-\frac{6\sqrt{2}-12}{2}$$

Since \frac{2\left(3\sqrt{2}+3\right)}{2} and \frac{6\sqrt{2}-12}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(3\sqrt{2}+3\right)-\left(6\sqrt{2}-12\right)}{2}$$

Do the multiplications in 2\left(3\sqrt{2}+3\right)-\left(6\sqrt{2}-12\right).

$$\frac{6\sqrt{2}+6-6\sqrt{2}+12}{2}$$

Do the calculations in 6\sqrt{2}+6-6\sqrt{2}+12.

$$\frac{18}{2}$$

Factor

$$3^{2}$$