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

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

Evaluate

$$-\sqrt{2}-1\approx -2.414213562$$

Solution Steps

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

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

Consider \left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\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}.

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

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

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

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Cancel out 2 and 2.

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

To find the opposite of \sqrt{3}+1, find the opposite of each term.

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

Combine \sqrt{3} and -\sqrt{3} to get 0.

$$-\sqrt{2}-1$$