Solve sqrt{12}+1/sqrt{3-sqrt{2}}-sqrt{6}*sqrt{3}

$$\sqrt { 12 } + \frac { 1 } { \sqrt { 3 } - \sqrt { 2 } } - \sqrt { 6 } \times \sqrt { 3 }$$

Evaluate

$$3\sqrt{3}-2\sqrt{2}\approx 2.367725298$$

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}.

$$2\sqrt{3}+\frac{1}{\sqrt{3}-\sqrt{2}}-\sqrt{6}\sqrt{3}$$

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

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

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}.

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

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

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

Subtract 2 from 3 to get 1.

$$2\sqrt{3}+\frac{\sqrt{3}+\sqrt{2}}{1}-\sqrt{6}\sqrt{3}$$

Anything divided by one gives itself.

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

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

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

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}.

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

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

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

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

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