Solve 7-4sqrt{3}+1/7-4sqrt{3}

$$7-4 \sqrt{ 3 } + \frac{ 1 }{ 7-4 \sqrt{ 3 } }$$

Evaluate

$$14$$

Solution Steps

Rationalize the denominator of \frac{1}{7-4\sqrt{3}} by multiplying numerator and denominator by 7+4\sqrt{3}.

$$7-4\sqrt{3}+\frac{7+4\sqrt{3}}{\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)}$$

Consider \left(7-4\sqrt{3}\right)\left(7+4\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}.

$$7-4\sqrt{3}+\frac{7+4\sqrt{3}}{7^{2}-\left(-4\sqrt{3}\right)^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$7-4\sqrt{3}+\frac{7+4\sqrt{3}}{49-\left(-4\sqrt{3}\right)^{2}}$$

Expand \left(-4\sqrt{3}\right)^{2}.

$$7-4\sqrt{3}+\frac{7+4\sqrt{3}}{49-\left(-4\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -4 to the power of 2 and get 16.

$$7-4\sqrt{3}+\frac{7+4\sqrt{3}}{49-16\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$7-4\sqrt{3}+\frac{7+4\sqrt{3}}{49-16\times 3}$$

Multiply 16 and 3 to get 48.

$$7-4\sqrt{3}+\frac{7+4\sqrt{3}}{49-48}$$

Subtract 48 from 49 to get 1.

$$7-4\sqrt{3}+\frac{7+4\sqrt{3}}{1}$$

Anything divided by one gives itself.

$$7-4\sqrt{3}+7+4\sqrt{3}$$

Add 7 and 7 to get 14.

$$14-4\sqrt{3}+4\sqrt{3}$$

Combine -4\sqrt{3} and 4\sqrt{3} to get 0.

$$14$$

Factor

$$2\times 7$$