Solve 2/sqrt{3-1}quadfrac{sqrt{2}}{sqrt{2}+1}quad3/2-sqrt{3}

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

Evaluate

$$18\sqrt{3}+30-9\sqrt{6}-15\sqrt{2}\approx 17.918303416$$

Solution Steps

Rationalize the denominator of \frac{2}{\sqrt{3}-1} by multiplying numerator and denominator by \sqrt{3}+1.

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

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Cancel out 2 and 2.

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

Rationalize the denominator of \frac{\sqrt{2}}{\sqrt{2}+1} by multiplying numerator and denominator by \sqrt{2}-1.

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

Consider \left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

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

Use the distributive property to multiply \sqrt{2} by \sqrt{2}-1.

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

The square of \sqrt{2} is 2.

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

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

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

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}.

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

Square 2. Square \sqrt{3}.

$$\left(\sqrt{3}+1\right)\left(2-\sqrt{2}\right)\times \frac{3\left(2+\sqrt{3}\right)}{4-3}$$

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

Apply the distributive property by multiplying each term of \sqrt{3}+1 by each term of 2-\sqrt{2}.

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

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

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

Apply the distributive property by multiplying each term of 2\sqrt{3}-\sqrt{6}+2-\sqrt{2} by each term of 6+3\sqrt{3}.

$$12\sqrt{3}+6\left(\sqrt{3}\right)^{2}-6\sqrt{6}-3\sqrt{3}\sqrt{6}+12+6\sqrt{3}-6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

The square of \sqrt{3} is 3.

$$12\sqrt{3}+6\times 3-6\sqrt{6}-3\sqrt{3}\sqrt{6}+12+6\sqrt{3}-6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

Multiply 6 and 3 to get 18.

$$12\sqrt{3}+18-6\sqrt{6}-3\sqrt{3}\sqrt{6}+12+6\sqrt{3}-6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$12\sqrt{3}+18-6\sqrt{6}-3\sqrt{3}\sqrt{3}\sqrt{2}+12+6\sqrt{3}-6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$12\sqrt{3}+18-6\sqrt{6}-3\times 3\sqrt{2}+12+6\sqrt{3}-6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

Multiply -3 and 3 to get -9.

$$12\sqrt{3}+18-6\sqrt{6}-9\sqrt{2}+12+6\sqrt{3}-6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

Add 18 and 12 to get 30.

$$12\sqrt{3}+30-6\sqrt{6}-9\sqrt{2}+6\sqrt{3}-6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

Combine 12\sqrt{3} and 6\sqrt{3} to get 18\sqrt{3}.

$$18\sqrt{3}+30-6\sqrt{6}-9\sqrt{2}-6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

Combine -9\sqrt{2} and -6\sqrt{2} to get -15\sqrt{2}.

$$18\sqrt{3}+30-6\sqrt{6}-15\sqrt{2}-3\sqrt{3}\sqrt{2}$$

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

$$18\sqrt{3}+30-6\sqrt{6}-15\sqrt{2}-3\sqrt{6}$$

Combine -6\sqrt{6} and -3\sqrt{6} to get -9\sqrt{6}.

$$18\sqrt{3}+30-9\sqrt{6}-15\sqrt{2}$$