Solve 3/sqrt{5-sqrt{2}}+5/sqrt{7+sqrt{2}}-2/sqrt{7-sqrt{5}}

$$\frac { 3 } { \sqrt { 5 } - \sqrt { 2 } } + \frac { 5 } { \sqrt { 7 } + \sqrt { 2 } } - \frac { 2 } { \sqrt { 7 } - \sqrt { 5 } }$$

Evaluate

$$0$$

Solution Steps

Rationalize the denominator of \frac{3}{\sqrt{5}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{5}+\sqrt{2}.

$$\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\frac{5}{\sqrt{7}+\sqrt{2}}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Consider \left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\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}.

$$\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}}+\frac{5}{\sqrt{7}+\sqrt{2}}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Square \sqrt{5}. Square \sqrt{2}.

$$\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\frac{5}{\sqrt{7}+\sqrt{2}}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Subtract 2 from 5 to get 3.

$$\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\frac{5}{\sqrt{7}+\sqrt{2}}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Cancel out 3 and 3.

$$\sqrt{5}+\sqrt{2}+\frac{5}{\sqrt{7}+\sqrt{2}}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Rationalize the denominator of \frac{5}{\sqrt{7}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{7}-\sqrt{2}.

$$\sqrt{5}+\sqrt{2}+\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Consider \left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\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}.

$$\sqrt{5}+\sqrt{2}+\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{2}\right)^{2}}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Square \sqrt{7}. Square \sqrt{2}.

$$\sqrt{5}+\sqrt{2}+\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Subtract 2 from 7 to get 5.

$$\sqrt{5}+\sqrt{2}+\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{5}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Cancel out 5 and 5.

$$\sqrt{5}+\sqrt{2}+\sqrt{7}-\sqrt{2}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

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

$$\sqrt{5}+\sqrt{7}-\frac{2}{\sqrt{7}-\sqrt{5}}$$

Rationalize the denominator of \frac{2}{\sqrt{7}-\sqrt{5}} by multiplying numerator and denominator by \sqrt{7}+\sqrt{5}.

$$\sqrt{5}+\sqrt{7}-\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}$$

Consider \left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\sqrt{5}+\sqrt{7}-\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{5}\right)^{2}}$$

Square \sqrt{7}. Square \sqrt{5}.

$$\sqrt{5}+\sqrt{7}-\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{7-5}$$

Subtract 5 from 7 to get 2.

$$\sqrt{5}+\sqrt{7}-\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{2}$$

Cancel out 2 and 2.

$$\sqrt{5}+\sqrt{7}-\left(\sqrt{7}+\sqrt{5}\right)$$

To find the opposite of \sqrt{7}+\sqrt{5}, find the opposite of each term.

$$\sqrt{5}+\sqrt{7}-\sqrt{7}-\sqrt{5}$$

Combine \sqrt{7} and -\sqrt{7} to get 0.

$$\sqrt{5}-\sqrt{5}$$

Combine \sqrt{5} and -\sqrt{5} to get 0.

$$0$$

Factor

$$0$$