Solve sqrt{27}-sqrt{2}*sqrt{6}-sqrt{18}-1/2+sqrt{3}

$$\sqrt{ 27 } - \sqrt{ 2 } \times \sqrt{ 6 } - \sqrt{ 18 } - \frac{ 1 }{ 2+ \sqrt{ 3 } }$$

Evaluate

$$2\sqrt{3}-3\sqrt{2}-2\approx -2.778539072$$

Solution Steps

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}.

$$3\sqrt{3}-\sqrt{2}\sqrt{6}-\sqrt{18}-\frac{1}{2+\sqrt{3}}$$

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

$$3\sqrt{3}-\sqrt{2}\sqrt{2}\sqrt{3}-\sqrt{18}-\frac{1}{2+\sqrt{3}}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$3\sqrt{3}-2\sqrt{3}-\sqrt{18}-\frac{1}{2+\sqrt{3}}$$

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

$$\sqrt{3}-\sqrt{18}-\frac{1}{2+\sqrt{3}}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

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

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

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

The opposite of -\sqrt{3} is \sqrt{3}.

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

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

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