Solve frac{2-sqrt{2}}{sqrt{2}-1}+sqrt{8}-sqrt{18}

$$\frac{ 2- \sqrt{ 2 } }{ \sqrt{ 2 } -1 } + \sqrt{ 8 } - \sqrt{ 18 }$$

Evaluate

$$0$$

Solution Steps

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

$$\frac{\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{8}-\sqrt{18}$$

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}.

$$\frac{\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}\right)^{2}-1^{2}}+\sqrt{8}-\sqrt{18}$$

Square \sqrt{2}. Square 1.

$$\frac{\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)}{2-1}+\sqrt{8}-\sqrt{18}$$

Subtract 1 from 2 to get 1.

$$\frac{\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)}{1}+\sqrt{8}-\sqrt{18}$$

Anything divided by one gives itself.

$$\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)+\sqrt{8}-\sqrt{18}$$

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}.

$$\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)+2\sqrt{2}-\sqrt{18}$$

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}.

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

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

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

Apply the distributive property by multiplying each term of 2-\sqrt{2} by each term of \sqrt{2}+1.

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

The square of \sqrt{2} is 2.

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

Subtract 2 from 2 to get 0.

$$2\sqrt{2}-\sqrt{2}-\sqrt{2}$$

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

$$\sqrt{2}-\sqrt{2}$$

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

$$0$$

Factor

$$0$$