If \(\frac{\sqrt{38-5\sqrt{3} } }{\sqrt{26+7\sqrt{3} } }= \frac{a+b\sqrt{3} }{23} \) , b > 0, then the value of (b – a) is:

If \(\frac {8 + 2\sqrt 3}{3\sqrt 3 + 5} = a\sqrt 3 - b,\)  then the value of a + b is equal to:

If \(\frac{{22\sqrt 2 }}{{4\sqrt 2 - \sqrt {3\, + \,\sqrt 5 }}}\)  =  \(a + \sqrt 5 b\) , with a, b > 0, then what is the value of (ab) ∶ (a + b)?

The value of \(\frac{1}{{4 - \sqrt {15} }} - \frac{1}{{\sqrt {15} - \sqrt {14} }} + \frac{1}{{\sqrt {14} - \sqrt {13} }} - \frac{1}{{\sqrt {13} - \sqrt {12} }} + \frac{1}{{\sqrt {12} - \sqrt {11} }} - \frac{1}{{\sqrt {11} - \sqrt {10} }} + \frac{1}{{\sqrt {10} - 3}} - \frac{1}{{3 - \sqrt 8 }}\)  is:

If x = \(\sqrt {1 + \frac{{\sqrt 3 }}{2}} - \sqrt {1 - \frac{{{\kern 1pt} \sqrt 3 }}{2}} \) , then the value of  \(\frac{{\sqrt 3 - x}}{{\sqrt 3 + x}}\)  (corrected to two decimal places) is:

Sum of four times a fraction and 7 times its reciprocal is 16. What is the fraction?

Let \(x\; = \;\sqrt[6]{{27}} - \sqrt {6\frac{3}{4}}\) and \(\;y\; = \;\frac{{\sqrt {45\;} \; + \;\sqrt {605} \; + \;\sqrt {245} }}{{\sqrt {80\;} \; + \;\sqrt {125} }}\) , then the value of x ²+ y ²is :

The value of \(\frac{{2\sqrt {10} }}{{\sqrt 5 \; + \;\sqrt 2 - \sqrt 7 }} - \sqrt {\frac{{\sqrt 5 - 2}}{{\sqrt 5 \; + \;2}}} - \frac{3}{{\sqrt 7 - 2}}\;\) is:

Which of the following is TRUE?

\({\rm{I}}.{\rm{\;}}\frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt 5 }}\)

\({\rm{II}}.\;\frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt 5 }}\)

\({\rm{III}}.\;\frac{1}{{\sqrt 5 }} > \frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt[4]{{29}}}}\)

\({\rm{IV}}.{\rm{\;}}\frac{1}{{\sqrt 5 }} > \frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt[3]{{12}}}}\)

Which of the following number is irrational?

Which of the following numbers will have an irrational square root?

What is the square root of 16 + 6√7?

A non-terminating but recurring decimal is: