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

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

Evaluate

$$1$$

Solution Steps

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

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

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

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

Square 2. Square \sqrt{3}.

$$\frac{\frac{3\left(2-\sqrt{3}\right)}{4-3}-\frac{2}{2-\sqrt{3}}}{2-5\sqrt{3}}$$

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

Square 2. Square \sqrt{3}.

$$\frac{3\left(2-\sqrt{3}\right)-\frac{2\left(2+\sqrt{3}\right)}{4-3}}{2-5\sqrt{3}}$$

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

Consider \left(2-5\sqrt{3}\right)\left(2+5\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}.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3\left(2-\sqrt{3}\right)-2\left(2+\sqrt{3}\right)\right)\left(2+5\sqrt{3}\right)}{4-\left(-5\sqrt{3}\right)^{2}}$$

Expand \left(-5\sqrt{3}\right)^{2}.

$$\frac{\left(3\left(2-\sqrt{3}\right)-2\left(2+\sqrt{3}\right)\right)\left(2+5\sqrt{3}\right)}{4-\left(-5\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -5 to the power of 2 and get 25.

$$\frac{\left(3\left(2-\sqrt{3}\right)-2\left(2+\sqrt{3}\right)\right)\left(2+5\sqrt{3}\right)}{4-25\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(3\left(2-\sqrt{3}\right)-2\left(2+\sqrt{3}\right)\right)\left(2+5\sqrt{3}\right)}{4-25\times 3}$$

Multiply 25 and 3 to get 75.

$$\frac{\left(3\left(2-\sqrt{3}\right)-2\left(2+\sqrt{3}\right)\right)\left(2+5\sqrt{3}\right)}{4-75}$$

Subtract 75 from 4 to get -71.

$$\frac{\left(3\left(2-\sqrt{3}\right)-2\left(2+\sqrt{3}\right)\right)\left(2+5\sqrt{3}\right)}{-71}$$

Use the distributive property to multiply 3 by 2-\sqrt{3}.

$$\frac{\left(6-3\sqrt{3}-2\left(2+\sqrt{3}\right)\right)\left(2+5\sqrt{3}\right)}{-71}$$

Use the distributive property to multiply 2 by 2+\sqrt{3}.

$$\frac{\left(6-3\sqrt{3}-\left(4+2\sqrt{3}\right)\right)\left(2+5\sqrt{3}\right)}{-71}$$

To find the opposite of 4+2\sqrt{3}, find the opposite of each term.

$$\frac{\left(6-3\sqrt{3}-4-2\sqrt{3}\right)\left(2+5\sqrt{3}\right)}{-71}$$

Subtract 4 from 6 to get 2.

$$\frac{\left(2-3\sqrt{3}-2\sqrt{3}\right)\left(2+5\sqrt{3}\right)}{-71}$$

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

$$\frac{\left(2-5\sqrt{3}\right)\left(2+5\sqrt{3}\right)}{-71}$$

Consider \left(2-5\sqrt{3}\right)\left(2+5\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}.

$$\frac{2^{2}-\left(5\sqrt{3}\right)^{2}}{-71}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4-\left(5\sqrt{3}\right)^{2}}{-71}$$

Expand \left(5\sqrt{3}\right)^{2}.

$$\frac{4-5^{2}\left(\sqrt{3}\right)^{2}}{-71}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{4-25\left(\sqrt{3}\right)^{2}}{-71}$$

The square of \sqrt{3} is 3.

$$\frac{4-25\times 3}{-71}$$

Multiply 25 and 3 to get 75.

$$\frac{4-75}{-71}$$

Subtract 75 from 4 to get -71.

$$\frac{-71}{-71}$$

Divide -71 by -71 to get 1.

$$1$$

Factor

$$1$$