Solve sqrt{27}+sqrt{3div25}-1/sqrt{2-1}

$$\sqrt{ 27 } + \sqrt{ 3 \div 25 } - \frac{ 1 }{ \sqrt{ 2 } -1 }$$

Evaluate

$$\frac{16\sqrt{3}}{5}-\sqrt{2}-1\approx 3.128349022$$

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}.

$$3\sqrt{3}+\sqrt{\frac{3}{25}}-\frac{1}{\sqrt{2}-1}$$

Rewrite the square root of the division \sqrt{\frac{3}{25}} as the division of square roots \frac{\sqrt{3}}{\sqrt{25}}.

$$3\sqrt{3}+\frac{\sqrt{3}}{\sqrt{25}}-\frac{1}{\sqrt{2}-1}$$

Calculate the square root of 25 and get 5.

$$3\sqrt{3}+\frac{\sqrt{3}}{5}-\frac{1}{\sqrt{2}-1}$$

Combine 3\sqrt{3} and \frac{\sqrt{3}}{5} to get \frac{16}{5}\sqrt{3}.

$$\frac{16}{5}\sqrt{3}-\frac{1}{\sqrt{2}-1}$$

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

$$\frac{16}{5}\sqrt{3}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}$$

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

$$\frac{16}{5}\sqrt{3}-\frac{\sqrt{2}+1}{\left(\sqrt{2}\right)^{2}-1^{2}}$$

Square \sqrt{2}. Square 1.

$$\frac{16}{5}\sqrt{3}-\frac{\sqrt{2}+1}{2-1}$$

Subtract 1 from 2 to get 1.

$$\frac{16}{5}\sqrt{3}-\frac{\sqrt{2}+1}{1}$$

Anything divided by one gives itself.

$$\frac{16}{5}\sqrt{3}-\left(\sqrt{2}+1\right)$$

To find the opposite of \sqrt{2}+1, find the opposite of each term.

$$\frac{16}{5}\sqrt{3}-\sqrt{2}-1$$