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

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

Evaluate

$$\sqrt{6}+\sqrt{10}-\sqrt{15}-2\approx -0.261215943$$

Solution Steps

Divide \sqrt{2}-\sqrt{5} by \sqrt{2}-\sqrt{5} to get 1.

$$\frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}+\sqrt{3}}\times 1$$

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

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

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

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

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

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

Subtract 3 from 2 to get -1.

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

Anything divided by -1 gives its opposite.

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

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

$$\left(-\left(\left(\sqrt{2}\right)^{2}-\sqrt{2}\sqrt{3}-\sqrt{5}\sqrt{2}+\sqrt{3}\sqrt{5}\right)\right)\times 1$$

The square of \sqrt{2} is 2.

$$\left(-\left(2-\sqrt{2}\sqrt{3}-\sqrt{5}\sqrt{2}+\sqrt{3}\sqrt{5}\right)\right)\times 1$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

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

$$\left(-\left(2-\sqrt{6}-\sqrt{10}+\sqrt{3}\sqrt{5}\right)\right)\times 1$$

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

$$\left(-\left(2-\sqrt{6}-\sqrt{10}+\sqrt{15}\right)\right)\times 1$$

To find the opposite of 2-\sqrt{6}-\sqrt{10}+\sqrt{15}, find the opposite of each term.

$$\left(-2-\left(-\sqrt{6}\right)-\left(-\sqrt{10}\right)-\sqrt{15}\right)\times 1$$

The opposite of -\sqrt{6} is \sqrt{6}.

$$\left(-2+\sqrt{6}-\left(-\sqrt{10}\right)-\sqrt{15}\right)\times 1$$

The opposite of -\sqrt{10} is \sqrt{10}.

$$\left(-2+\sqrt{6}+\sqrt{10}-\sqrt{15}\right)\times 1$$

Use the distributive property to multiply -2+\sqrt{6}+\sqrt{10}-\sqrt{15} by 1.

$$-2+\sqrt{6}+\sqrt{10}-\sqrt{15}$$

Factor

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