Solve frac{sqrt{3}-sqrt{2}}{sqrt{12}-sqrt{8}}-frac{sqrt{27}}{3}-3/2

$$\frac{ \sqrt{ 3 } - \sqrt{ 2 } }{ \sqrt{ 12 } - \sqrt{ 8 } } - \frac{ \sqrt{ 27 } }{ 3 } - \frac{ 3 }{ 2 }$$

Evaluate

$$-\sqrt{3}-1\approx -2.732050808$$

Short Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}-\sqrt{8}}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}-2\sqrt{2}}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

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

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{\left(2\sqrt{3}-2\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

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

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{\left(2\sqrt{3}\right)^{2}-\left(-2\sqrt{2}\right)^{2}}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

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

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{2^{2}\left(\sqrt{3}\right)^{2}-\left(-2\sqrt{2}\right)^{2}}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{4\left(\sqrt{3}\right)^{2}-\left(-2\sqrt{2}\right)^{2}}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

The square of \sqrt{3} is 3.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{4\times 3-\left(-2\sqrt{2}\right)^{2}}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

Multiply 4 and 3 to get 12.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{12-\left(-2\sqrt{2}\right)^{2}}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

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

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{12-\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{12-4\left(\sqrt{2}\right)^{2}}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{12-4\times 2}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

Multiply 4 and 2 to get 8.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{12-8}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

Subtract 8 from 12 to get 4.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{4}-\frac{\sqrt{27}}{3}-\frac{3}{2}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

Cancel out 3 and 3.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{3}{2} times \frac{2}{2}.

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

Since \frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{4} and \frac{3\times 2}{4} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+2\sqrt{2}\right)-3\times 2.

$$\frac{6+2\sqrt{6}-2\sqrt{6}-4-6}{4}-\sqrt{3}$$

Do the calculations in 6+2\sqrt{6}-2\sqrt{6}-4-6.

$$\frac{-4}{4}-\sqrt{3}$$

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

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

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

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

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

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

$$\frac{6-2\left(\sqrt{2}\right)^{2}}{4}-\sqrt{3}-\frac{3}{2}$$

The square of \sqrt{2} is 2.

$$\frac{6-2\times 2}{4}-\sqrt{3}-\frac{3}{2}$$

Multiply -2 and 2 to get -4.

$$\frac{6-4}{4}-\sqrt{3}-\frac{3}{2}$$

Subtract 4 from 6 to get 2.

$$\frac{2}{4}-\sqrt{3}-\frac{3}{2}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{1}{2}-\sqrt{3}-\frac{3}{2}$$

Since \frac{1}{2} and \frac{3}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-3}{2}-\sqrt{3}$$

Subtract 3 from 1 to get -2.

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

Divide -2 by 2 to get -1.

$$-1-\sqrt{3}$$

Factor

$$-\sqrt{3}-1$$