The value of -6 - 2-5 -1 is -

We know that, √(6 + 2√(5)) 1 = nbsp;√(5+ 1 + 2√(5)) 1 = nbsp;√((√(5))^2+ 1^2 + 2√(5)) 1 = nbsp;√((√(5)+1)^2) 1 = √(5)+1 1 = √(5) nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Expansion of (a + b)^2

The value of ...

Question

The value of $\sqrt{6 + 2 \sqrt{5}} - 1$ is .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

\frac{1}{3 - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2}

\frac{1}{3 - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2}

x = 5 + 2 \sqrt{6}

y = 5 - 2 \sqrt{6}

\frac{1}{x^{2}} + \frac{1}{y^{2}}

\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}

\frac{1}{\sqrt{9} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{1}}

Join BYJU'S Learning Program