Solve 1/sqrt{3/2x+x/2}

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

Evaluate

$$\frac{2x}{x^{2}+\sqrt{3}}$$

Short Solution Steps

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

$$\frac{1}{\frac{\sqrt{3}}{2x}+\frac{xx}{2x}}$$

Since \frac{\sqrt{3}}{2x} and \frac{xx}{2x} have the same denominator, add them by adding their numerators.

$$\frac{1}{\frac{\sqrt{3}+xx}{2x}}$$

Do the multiplications in \sqrt{3}+xx.

$$\frac{1}{\frac{\sqrt{3}+x^{2}}{2x}}$$

Divide 1 by \frac{\sqrt{3}+x^{2}}{2x} by multiplying 1 by the reciprocal of \frac{\sqrt{3}+x^{2}}{2x}.

$$\frac{2x}{\sqrt{3}+x^{2}}$$

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

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

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

The square of \sqrt{3} is 3.

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

Use the distributive property to multiply 2x by \sqrt{3}-x^{2}.

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

Factor

$$\frac{2x}{x^{2}+\sqrt{3}}$$