Solve frac{2sqrt{12}+sqrt{3}}{sqrt{3}}-(1-sqrt{3})(sqrt{3}+1)+(pi-2020)^0

$$\frac { 2 \sqrt { 12 } + \sqrt { 3 } } { \sqrt { 3 } } - ( 1 - \sqrt { 3 } ) ( \sqrt { 3 } + 1 ) + ( \pi - 2020 ) ^ { 0 }$$

Evaluate

$$8$$

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}.

$$\frac{2\times 2\sqrt{3}+\sqrt{3}}{\sqrt{3}}-\left(1-\sqrt{3}\right)\left(\sqrt{3}+1\right)+\left(\pi -2020\right)^{0}$$

Multiply 2 and 2 to get 4.

$$\frac{4\sqrt{3}+\sqrt{3}}{\sqrt{3}}-\left(1-\sqrt{3}\right)\left(\sqrt{3}+1\right)+\left(\pi -2020\right)^{0}$$

Combine 4\sqrt{3} and \sqrt{3} to get 5\sqrt{3}.

$$\frac{5\sqrt{3}}{\sqrt{3}}-\left(1-\sqrt{3}\right)\left(\sqrt{3}+1\right)+\left(\pi -2020\right)^{0}$$

Rationalize the denominator of \frac{5\sqrt{3}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{5\sqrt{3}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\left(1-\sqrt{3}\right)\left(\sqrt{3}+1\right)+\left(\pi -2020\right)^{0}$$

The square of \sqrt{3} is 3.

$$\frac{5\sqrt{3}\sqrt{3}}{3}-\left(1-\sqrt{3}\right)\left(\sqrt{3}+1\right)+\left(\pi -2020\right)^{0}$$

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

$$\frac{5\times 3}{3}-\left(1-\sqrt{3}\right)\left(\sqrt{3}+1\right)+\left(\pi -2020\right)^{0}$$

Multiply 5 and 3 to get 15.

$$\frac{15}{3}-\left(1-\sqrt{3}\right)\left(\sqrt{3}+1\right)+\left(\pi -2020\right)^{0}$$

Divide 15 by 3 to get 5.

$$5-\left(1-\sqrt{3}\right)\left(\sqrt{3}+1\right)+\left(\pi -2020\right)^{0}$$

Consider \left(1-\sqrt{3}\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}. Square 1.

$$5-\left(1-\left(\sqrt{3}\right)^{2}\right)+\left(\pi -2020\right)^{0}$$

The square of \sqrt{3} is 3.

$$5-\left(1-3\right)+\left(\pi -2020\right)^{0}$$

Subtract 3 from 1 to get -2.

$$5-\left(-2\right)+\left(\pi -2020\right)^{0}$$

The opposite of -2 is 2.

$$5+2+\left(\pi -2020\right)^{0}$$

Add 5 and 2 to get 7.

$$7+\left(\pi -2020\right)^{0}$$

Calculate \pi -2020 to the power of 0 and get 1.

$$7+1$$

Add 7 and 1 to get 8.

$$8$$

Factor

$$2^{3}$$