Solve frac{sqrt{2}-sqrt{3}}{sqrt{5}+sqrt{6}}

$$\frac { \sqrt { 2 } - \sqrt { 3 } } { \sqrt { 5 } + \sqrt { 6 } }$$

Evaluate

$$\sqrt{15}+2\sqrt{3}-\sqrt{10}-3\sqrt{2}\approx -0.067833386$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{2}-\sqrt{3}}{\sqrt{5}+\sqrt{6}} by multiplying numerator and denominator by \sqrt{5}-\sqrt{6}.

$$\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{6}\right)}{\left(\sqrt{5}+\sqrt{6}\right)\left(\sqrt{5}-\sqrt{6}\right)}$$

Consider \left(\sqrt{5}+\sqrt{6}\right)\left(\sqrt{5}-\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}.

$$\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{6}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{6}\right)^{2}}$$

Square \sqrt{5}. Square \sqrt{6}.

$$\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{6}\right)}{5-6}$$

Subtract 6 from 5 to get -1.

$$\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{6}\right)}{-1}$$

Anything divided by -1 gives its opposite.

$$-\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{6}\right)$$

Apply the distributive property by multiplying each term of \sqrt{2}-\sqrt{3} by each term of \sqrt{5}-\sqrt{6}.

$$-\left(\sqrt{2}\sqrt{5}-\sqrt{2}\sqrt{6}-\sqrt{3}\sqrt{5}+\sqrt{3}\sqrt{6}\right)$$

To multiply \sqrt{2} and \sqrt{5}, multiply the numbers under the square root.

$$-\left(\sqrt{10}-\sqrt{2}\sqrt{6}-\sqrt{3}\sqrt{5}+\sqrt{3}\sqrt{6}\right)$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$-\left(\sqrt{10}-\sqrt{2}\sqrt{2}\sqrt{3}-\sqrt{3}\sqrt{5}+\sqrt{3}\sqrt{6}\right)$$

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

$$-\left(\sqrt{10}-2\sqrt{3}-\sqrt{3}\sqrt{5}+\sqrt{3}\sqrt{6}\right)$$

To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.

$$-\left(\sqrt{10}-2\sqrt{3}-\sqrt{15}+\sqrt{3}\sqrt{6}\right)$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$-\left(\sqrt{10}-2\sqrt{3}-\sqrt{15}+\sqrt{3}\sqrt{3}\sqrt{2}\right)$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$-\left(\sqrt{10}-2\sqrt{3}-\sqrt{15}+3\sqrt{2}\right)$$

To find the opposite of \sqrt{10}-2\sqrt{3}-\sqrt{15}+3\sqrt{2}, find the opposite of each term.

$$-\sqrt{10}-\left(-2\sqrt{3}\right)-\left(-\sqrt{15}\right)-3\sqrt{2}$$

The opposite of -2\sqrt{3} is 2\sqrt{3}.

$$-\sqrt{10}+2\sqrt{3}-\left(-\sqrt{15}\right)-3\sqrt{2}$$

The opposite of -\sqrt{15} is \sqrt{15}.

$$-\sqrt{10}+2\sqrt{3}+\sqrt{15}-3\sqrt{2}$$