Solve H=frac{sqrt{3}+sqrt{2}}{sqrt{3}-sqrt{2}}+frac{sqrt{3}-sqrt{2}}{sqrt{3}+sqrt{2}}

$$H = \frac { \sqrt { 3 } + \sqrt { 2 } } { \sqrt { 3 } - \sqrt { 2 } } + \frac { \sqrt { 3 } - \sqrt { 2 } } { \sqrt { 3 } + \sqrt { 2 } }$$

Solve for H

$$H=10$$

Steps for Solving Linear Equation

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

$$H=\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$

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

$$H=\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$

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

$$H=\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{3-2}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$

Subtract 2 from 3 to get 1.

$$H=\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{1}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$

Anything divided by one gives itself.

$$H=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$

Multiply \sqrt{3}+\sqrt{2} and \sqrt{3}+\sqrt{2} to get \left(\sqrt{3}+\sqrt{2}\right)^{2}.

$$H=\left(\sqrt{3}+\sqrt{2}\right)^{2}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$

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

$$H=\left(\sqrt{3}+\sqrt{2}\right)^{2}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}$$

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

$$H=\left(\sqrt{3}+\sqrt{2}\right)^{2}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

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

$$H=\left(\sqrt{3}+\sqrt{2}\right)^{2}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{3-2}$$

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

$$H=\left(\sqrt{3}+\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)$$

Multiply \sqrt{3}-\sqrt{2} and \sqrt{3}-\sqrt{2} to get \left(\sqrt{3}-\sqrt{2}\right)^{2}.

$$H=\left(\sqrt{3}+\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+\sqrt{2}\right)^{2}.

$$H=\left(\sqrt{3}\right)^{2}+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$H=3+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

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

$$H=3+2\sqrt{6}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$H=3+2\sqrt{6}+2+\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Add 3 and 2 to get 5.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-\sqrt{2}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

$$H=5+2\sqrt{6}+3-2\sqrt{6}+2$$

Add 3 and 2 to get 5.

$$H=5+2\sqrt{6}+5-2\sqrt{6}$$

Add 5 and 5 to get 10.

$$H=10+2\sqrt{6}-2\sqrt{6}$$

Combine 2\sqrt{6} and -2\sqrt{6} to get 0.

$$H=10$$

Assign H

$$H≔10$$