The question asks to simplify the expression \((\sqrt{3} + \sqrt{2})^2\). We can solve this using the algebraic identity for the square of a sum: \( (a + b)^2 = a^2 + 2ab + b^2 \)
Let \(a = \sqrt{3}\) and \(b = \sqrt{2}\). Substitute these values into the identity:
Add the results obtained:
\( (\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 \)
Combine the constant terms:
\( (3 + 2) + 2\sqrt{6} = 5 + 2\sqrt{6} \)
The simplified form of the expression is \(5 + 2\sqrt{6}\).
Rationalize:
$\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$
Simplify: \(\sqrt{7+4\sqrt{3}}\)
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:
\(\sqrt{46.24}+\sqrt{0.4624}+\sqrt{0.004624}\)