Simplify: \(\sqrt{12 + 6\sqrt{3}}\)
\(3 + \sqrt{3}\)
Assume \(\sqrt{12 + 6\sqrt{3}} = a + b\sqrt{3}\) for rational a, b.
Squaring: \((a + b\sqrt{3})^2 = a^2 + 3b^2 + 2ab\sqrt{3}\)
Equating with \(12 + 6\sqrt{3}\): \(a^2 + 3b^2 = 12\) and \(2ab = 6 \Rightarrow ab = 3\).
Try \(a = 3,\ b = 1\): \(9 + 3 = 12\) ✓ and \(3 \times 1 = 3\) ✓.
Hence \(\sqrt{12 + 6\sqrt{3}} = 3 + \sqrt{3}\).
Rationalize:
$\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$
The value of \(\frac{{{{\left( {251} \right)}^3} + {{\left( {249} \right)}^3}}}{{25.1 \times 25.1 - 624.99 + 24.9 \times 24.9}}\) is 5 × 10 k , where the value of k is :
Find the value of m in \(\left(\frac{2}{7}\right)^{-3} \times \left(\frac{2}{7}\right)^{-5}=\left (\frac{2}{7}\right)^{-3m+1}\)
If √625 = 25; then√(.00000625/25)is:
A. 0.0025
B. 0.001
C. 0.0001
D. 0.0005Find the value of:
\(\sqrt{150}-\sqrt{54}-\sqrt{24}\)
If \(\sqrt{4624}=68\) , then the value of:
\(\sqrt{46.24}+\sqrt{0.4624}+\sqrt{0.004624}\)