Solve (1 + 1/(sqrt(3)))/(1 - 1/(sqrt(3))) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Evaluate

$\sqrt{3}+2\approx 3.732050808$

Short Solution Steps

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

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

The square of $\sqrt{3}$ is $3$.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply $1$ times $\frac{3}{3}$.

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

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

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

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

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

The square of $\sqrt{3}$ is $3$.

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

Express $1\times \frac{\sqrt{3}}{3}$ as a single fraction.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply $1$ times $\frac{3}{3}$.

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

Since $\frac{3}{3}$ and $\frac{\sqrt{3}}{3}$ have the same denominator, subtract them by subtracting their numerators.

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

Divide $\frac{3+\sqrt{3}}{3}$ by $\frac{3-\sqrt{3}}{3}$ by multiplying $\frac{3+\sqrt{3}}{3}$ by the reciprocal of $\frac{3-\sqrt{3}}{3}$.

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

Cancel out $3$ in both numerator and denominator.

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

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

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

Consider $\left(-\sqrt{3}+3\right)\left(-\sqrt{3}-3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

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

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

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

Calculate $-1$ to the power of $2$ and get $1$.

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

The square of $\sqrt{3}$ is $3$.

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

Multiply $1$ and $3$ to get $3$.

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

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

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

Subtract $9$ from $3$ to get $-6$.

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

Apply the distributive property by multiplying each term of $\sqrt{3}+3$ by each term of $-\sqrt{3}-3$.

$$\frac{-\left(\sqrt{3}\right)^{2}-3\sqrt{3}-3\sqrt{3}-9}{-6}$$

The square of $\sqrt{3}$ is $3$.

$$\frac{-3-3\sqrt{3}-3\sqrt{3}-9}{-6}$$

Combine $-3\sqrt{3}$ and $-3\sqrt{3}$ to get $-6\sqrt{3}$.

$$\frac{-3-6\sqrt{3}-9}{-6}$$

Subtract $9$ from $-3$ to get $-12$.

$$\frac{-12-6\sqrt{3}}{-6}$$

Divide each term of $-12-6\sqrt{3}$ by $-6$ to get $2+\sqrt{3}$.

$$2+\sqrt{3}$$