Solve frac{3+2sqrt{2}}{2+sqrt{2}}*frac{sqrt{2}-1}{sqrt{2}}

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

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

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

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

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

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

Square 2. Square \sqrt{2}.

$$\frac{\left(3+2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{4-2}\times \frac{\sqrt{2}-1}{\sqrt{2}}$$

Subtract 2 from 4 to get 2.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply \frac{\left(3+2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2} times \frac{\left(\sqrt{2}-1\right)\sqrt{2}}{2} by multiplying numerator times numerator and denominator times denominator.

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

Multiply 2 and 2 to get 4.

$$\frac{\left(3+2\sqrt{2}\right)\left(2-\sqrt{2}\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

Apply the distributive property by multiplying each term of 3+2\sqrt{2} by each term of 2-\sqrt{2}.

$$\frac{\left(6-3\sqrt{2}+4\sqrt{2}-2\left(\sqrt{2}\right)^{2}\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

Combine -3\sqrt{2} and 4\sqrt{2} to get \sqrt{2}.

$$\frac{\left(6+\sqrt{2}-2\left(\sqrt{2}\right)^{2}\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

The square of \sqrt{2} is 2.

$$\frac{\left(6+\sqrt{2}-2\times 2\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

Multiply -2 and 2 to get -4.

$$\frac{\left(6+\sqrt{2}-4\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

Subtract 4 from 6 to get 2.

$$\frac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

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

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

The square of \sqrt{2} is 2.

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

Add -2 and 2 to get 0.

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

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

$$\frac{\sqrt{2}\sqrt{2}}{4}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$