Solve 4/sqrt{7+sqrt{3}}+3/sqrt{6-sqrt{3}}-1/sqrt{7-sqrt{6}}

$$\frac { 4 } { \sqrt { 7 } + \sqrt { 3 } } + \frac { 3 } { \sqrt { 6 } - \sqrt { 3 } } - \frac { 1 } { \sqrt { 7 } - \sqrt { 6 } }$$

Evaluate

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Solution Steps

Rationalize the denominator of \frac{4}{\sqrt{7}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{7}-\sqrt{3}.

$$\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}+\frac{3}{\sqrt{6}-\sqrt{3}}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

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

$$\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}+\frac{3}{\sqrt{6}-\sqrt{3}}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

Square \sqrt{7}. Square \sqrt{3}.

$$\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{7-3}+\frac{3}{\sqrt{6}-\sqrt{3}}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

Subtract 3 from 7 to get 4.

$$\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{4}+\frac{3}{\sqrt{6}-\sqrt{3}}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

Cancel out 4 and 4.

$$\sqrt{7}-\sqrt{3}+\frac{3}{\sqrt{6}-\sqrt{3}}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

Rationalize the denominator of \frac{3}{\sqrt{6}-\sqrt{3}} by multiplying numerator and denominator by \sqrt{6}+\sqrt{3}.

$$\sqrt{7}-\sqrt{3}+\frac{3\left(\sqrt{6}+\sqrt{3}\right)}{\left(\sqrt{6}-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{3}\right)}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

Consider \left(\sqrt{6}-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\sqrt{7}-\sqrt{3}+\frac{3\left(\sqrt{6}+\sqrt{3}\right)}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

Square \sqrt{6}. Square \sqrt{3}.

$$\sqrt{7}-\sqrt{3}+\frac{3\left(\sqrt{6}+\sqrt{3}\right)}{6-3}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

Subtract 3 from 6 to get 3.

$$\sqrt{7}-\sqrt{3}+\frac{3\left(\sqrt{6}+\sqrt{3}\right)}{3}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

Cancel out 3 and 3.

$$\sqrt{7}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

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

$$\sqrt{7}+\sqrt{6}-\frac{1}{\sqrt{7}-\sqrt{6}}$$

Rationalize the denominator of \frac{1}{\sqrt{7}-\sqrt{6}} by multiplying numerator and denominator by \sqrt{7}+\sqrt{6}.

$$\sqrt{7}+\sqrt{6}-\frac{\sqrt{7}+\sqrt{6}}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}$$

Consider \left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{7}. Square \sqrt{6}.

$$\sqrt{7}+\sqrt{6}-\frac{\sqrt{7}+\sqrt{6}}{7-6}$$

Subtract 6 from 7 to get 1.

$$\sqrt{7}+\sqrt{6}-\frac{\sqrt{7}+\sqrt{6}}{1}$$

Anything divided by one gives itself.

$$\sqrt{7}+\sqrt{6}-\left(\sqrt{7}+\sqrt{6}\right)$$

To find the opposite of \sqrt{7}+\sqrt{6}, find the opposite of each term.

$$\sqrt{7}+\sqrt{6}-\sqrt{7}-\sqrt{6}$$

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

$$\sqrt{6}-\sqrt{6}$$

Combine \sqrt{6} and -\sqrt{6} to get 0.

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Factor

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