Solve 5/sqrt{3-sqrt{5}}=5/sqrt{3-sqrt{5}}*frac{sqrt{3}+sqrt{5}}{sqrt{3}+sqrt{5}}=frac{5(sqrt{3}+sqrt{5})}{3-5}

$$\frac { 5 } { \sqrt { 3 } - \sqrt { 5 } } = \frac { 5 } { \sqrt { 3 } - \sqrt { 5 } } \times \frac { \sqrt { 3 } + \sqrt { 5 } } { \sqrt { 3 } + \sqrt { 5 } } = \frac { 5 ( \sqrt { 3 } + \sqrt { 5 } ) } { 3 - 5 }$$

Verify

Solution Steps

Divide \sqrt{3}+\sqrt{5} by \sqrt{3}+\sqrt{5} to get 1.

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

Divide \sqrt{3}+\sqrt{5} by \sqrt{3}+\sqrt{5} to get 1.

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

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

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

Consider \left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\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}.

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

Square \sqrt{3}. Square \sqrt{5}.

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

Subtract 5 from 3 to get -2.

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

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

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

Consider \left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\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}.

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

Square \sqrt{3}. Square \sqrt{5}.

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

Subtract 5 from 3 to get -2.

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

Express \frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}\times 1 as a single fraction.

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

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

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

Consider \left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\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}.

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

Square \sqrt{3}. Square \sqrt{5}.

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

Subtract 5 from 3 to get -2.

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

Express \frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}\times 1 as a single fraction.

$$\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}\text{ and }\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{3-5}$$

Subtract 5 from 3 to get -2.

$$\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}\text{ and }\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}$$

Use the distributive property to multiply 5 by \sqrt{3}+\sqrt{5}.

$$\frac{5\sqrt{3}+5\sqrt{5}}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}\text{ and }\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}$$

Use the distributive property to multiply 5 by \sqrt{3}+\sqrt{5}.

$$\frac{5\sqrt{3}+5\sqrt{5}}{-2}=\frac{5\sqrt{3}+5\sqrt{5}}{-2}\text{ and }\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}$$

Subtract \frac{5\sqrt{3}+5\sqrt{5}}{-2} from both sides.

$$\frac{5\sqrt{3}+5\sqrt{5}}{-2}-\frac{5\sqrt{3}+5\sqrt{5}}{-2}=0\text{ and }\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}$$

Combine \frac{5\sqrt{3}+5\sqrt{5}}{-2} and -\frac{5\sqrt{3}+5\sqrt{5}}{-2} to get 0.

$$0=0\text{ and }\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}$$

Compare 0 and 0.

$$\text{true}\text{ and }\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}$$

Use the distributive property to multiply 5 by \sqrt{3}+\sqrt{5}.

$$\text{true}\text{ and }\frac{5\sqrt{3}+5\sqrt{5}}{-2}=\frac{5\left(\sqrt{3}+\sqrt{5}\right)}{-2}$$

Use the distributive property to multiply 5 by \sqrt{3}+\sqrt{5}.

$$\text{true}\text{ and }\frac{5\sqrt{3}+5\sqrt{5}}{-2}=\frac{5\sqrt{3}+5\sqrt{5}}{-2}$$

Subtract \frac{5\sqrt{3}+5\sqrt{5}}{-2} from both sides.

$$\text{true}\text{ and }\frac{5\sqrt{3}+5\sqrt{5}}{-2}-\frac{5\sqrt{3}+5\sqrt{5}}{-2}=0$$

Combine \frac{5\sqrt{3}+5\sqrt{5}}{-2} and -\frac{5\sqrt{3}+5\sqrt{5}}{-2} to get 0.

$$\text{true}\text{ and }0=0$$

Compare 0 and 0.

$$\text{true}\text{ and }\text{true}$$

The conjunction of \text{true} and \text{true} is \text{true}.

$$\text{true}$$