Solve 3/3x-sqrt{27}

$$\frac { 3 } { 3 x - \sqrt { 27 } }$$

Evaluate

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

Short 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}.

$$\frac{3}{3x-3\sqrt{3}}$$

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

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

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

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

Expand \left(3x\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{3\left(3x+3\sqrt{3}\right)}{9x^{2}-\left(-3\sqrt{3}\right)^{2}}$$

Expand \left(-3\sqrt{3}\right)^{2}.

$$\frac{3\left(3x+3\sqrt{3}\right)}{9x^{2}-\left(-3\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -3 to the power of 2 and get 9.

$$\frac{3\left(3x+3\sqrt{3}\right)}{9x^{2}-9\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3\left(3x+3\sqrt{3}\right)}{9x^{2}-9\times 3}$$

Multiply 9 and 3 to get 27.

$$\frac{3\left(3x+3\sqrt{3}\right)}{9x^{2}-27}$$

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

$$\frac{9x+9\sqrt{3}}{9x^{2}-27}$$

Factor the expressions that are not already factored.

$$\frac{9\left(x+\sqrt{3}\right)}{9\left(x^{2}-3\right)}$$

Cancel out 9 in both numerator and denominator.

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

Differentiate w.r.t. x

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