Solve 3/4-sqrt{13}-6/sqrt{13-sqrt{7}}-2/3+sqrt{7}

$$\frac{ 3 }{ 4- \sqrt{ 13 } } - \frac{ 6 }{ \sqrt{ 13 } - \sqrt{ 7 } } - \frac{ 2 }{ 3+ \sqrt{ 7 } }$$

Evaluate

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Solution Steps

Rationalize the denominator of \frac{3}{4-\sqrt{13}} by multiplying numerator and denominator by 4+\sqrt{13}.

$$\frac{3\left(4+\sqrt{13}\right)}{\left(4-\sqrt{13}\right)\left(4+\sqrt{13}\right)}-\frac{6}{\sqrt{13}-\sqrt{7}}-\frac{2}{3+\sqrt{7}}$$

Consider \left(4-\sqrt{13}\right)\left(4+\sqrt{13}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{3\left(4+\sqrt{13}\right)}{4^{2}-\left(\sqrt{13}\right)^{2}}-\frac{6}{\sqrt{13}-\sqrt{7}}-\frac{2}{3+\sqrt{7}}$$

Square 4. Square \sqrt{13}.

$$\frac{3\left(4+\sqrt{13}\right)}{16-13}-\frac{6}{\sqrt{13}-\sqrt{7}}-\frac{2}{3+\sqrt{7}}$$

Subtract 13 from 16 to get 3.

$$\frac{3\left(4+\sqrt{13}\right)}{3}-\frac{6}{\sqrt{13}-\sqrt{7}}-\frac{2}{3+\sqrt{7}}$$

Cancel out 3 and 3.

$$4+\sqrt{13}-\frac{6}{\sqrt{13}-\sqrt{7}}-\frac{2}{3+\sqrt{7}}$$

Rationalize the denominator of \frac{6}{\sqrt{13}-\sqrt{7}} by multiplying numerator and denominator by \sqrt{13}+\sqrt{7}.

$$4+\sqrt{13}-\frac{6\left(\sqrt{13}+\sqrt{7}\right)}{\left(\sqrt{13}-\sqrt{7}\right)\left(\sqrt{13}+\sqrt{7}\right)}-\frac{2}{3+\sqrt{7}}$$

Consider \left(\sqrt{13}-\sqrt{7}\right)\left(\sqrt{13}+\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$4+\sqrt{13}-\frac{6\left(\sqrt{13}+\sqrt{7}\right)}{\left(\sqrt{13}\right)^{2}-\left(\sqrt{7}\right)^{2}}-\frac{2}{3+\sqrt{7}}$$

Square \sqrt{13}. Square \sqrt{7}.

$$4+\sqrt{13}-\frac{6\left(\sqrt{13}+\sqrt{7}\right)}{13-7}-\frac{2}{3+\sqrt{7}}$$

Subtract 7 from 13 to get 6.

$$4+\sqrt{13}-\frac{6\left(\sqrt{13}+\sqrt{7}\right)}{6}-\frac{2}{3+\sqrt{7}}$$

Cancel out 6 and 6.

$$4+\sqrt{13}-\left(\sqrt{13}+\sqrt{7}\right)-\frac{2}{3+\sqrt{7}}$$

To find the opposite of \sqrt{13}+\sqrt{7}, find the opposite of each term.

$$4+\sqrt{13}-\sqrt{13}-\sqrt{7}-\frac{2}{3+\sqrt{7}}$$

Combine \sqrt{13} and -\sqrt{13} to get 0.

$$4-\sqrt{7}-\frac{2}{3+\sqrt{7}}$$

Rationalize the denominator of \frac{2}{3+\sqrt{7}} by multiplying numerator and denominator by 3-\sqrt{7}.

$$4-\sqrt{7}-\frac{2\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}$$

Consider \left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$4-\sqrt{7}-\frac{2\left(3-\sqrt{7}\right)}{3^{2}-\left(\sqrt{7}\right)^{2}}$$

Square 3. Square \sqrt{7}.

$$4-\sqrt{7}-\frac{2\left(3-\sqrt{7}\right)}{9-7}$$

Subtract 7 from 9 to get 2.

$$4-\sqrt{7}-\frac{2\left(3-\sqrt{7}\right)}{2}$$

Cancel out 2 and 2.

$$4-\sqrt{7}-\left(3-\sqrt{7}\right)$$

To find the opposite of 3-\sqrt{7}, find the opposite of each term.

$$4-\sqrt{7}-3-\left(-\sqrt{7}\right)$$

The opposite of -\sqrt{7} is \sqrt{7}.

$$4-\sqrt{7}-3+\sqrt{7}$$

Subtract 3 from 4 to get 1.

$$1-\sqrt{7}+\sqrt{7}$$

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

$$1$$

Factor

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