Solve 3/sqrt{2-1}-6/sqrt{2}

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

Evaluate

$$3$$

Solution Steps

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

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

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}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

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

Rationalize the denominator of \frac{6}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Divide 6\sqrt{2} by 2 to get 3\sqrt{2}.

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

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

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

Subtract 3\sqrt{2} from 3\sqrt{2} to get 0.

$$3$$

Factor

$$3$$