Solve 19/sqrt{20+1}+6-2sqrt{5}

$$\frac { 19 } { \sqrt { 20 } + 1 } + 6 - 2 \sqrt { 5 }$$

Evaluate

$$5$$

Solution Steps

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

$$\frac{19}{2\sqrt{5}+1}+6-2\sqrt{5}$$

Rationalize the denominator of \frac{19}{2\sqrt{5}+1} by multiplying numerator and denominator by 2\sqrt{5}-1.

$$\frac{19\left(2\sqrt{5}-1\right)}{\left(2\sqrt{5}+1\right)\left(2\sqrt{5}-1\right)}+6-2\sqrt{5}$$

Consider \left(2\sqrt{5}+1\right)\left(2\sqrt{5}-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}.

$$\frac{19\left(2\sqrt{5}-1\right)}{\left(2\sqrt{5}\right)^{2}-1^{2}}+6-2\sqrt{5}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$\frac{19\left(2\sqrt{5}-1\right)}{2^{2}\left(\sqrt{5}\right)^{2}-1^{2}}+6-2\sqrt{5}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{19\left(2\sqrt{5}-1\right)}{4\left(\sqrt{5}\right)^{2}-1^{2}}+6-2\sqrt{5}$$

The square of \sqrt{5} is 5.

$$\frac{19\left(2\sqrt{5}-1\right)}{4\times 5-1^{2}}+6-2\sqrt{5}$$

Multiply 4 and 5 to get 20.

$$\frac{19\left(2\sqrt{5}-1\right)}{20-1^{2}}+6-2\sqrt{5}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{19\left(2\sqrt{5}-1\right)}{20-1}+6-2\sqrt{5}$$

Subtract 1 from 20 to get 19.

$$\frac{19\left(2\sqrt{5}-1\right)}{19}+6-2\sqrt{5}$$

Cancel out 19 and 19.

$$2\sqrt{5}-1+6-2\sqrt{5}$$

Add -1 and 6 to get 5.

$$2\sqrt{5}+5-2\sqrt{5}$$

Combine 2\sqrt{5} and -2\sqrt{5} to get 0.

$$5$$

Factor

$$5$$