Question

$\sqrt{a^2b^2x}$

Explanation:

Step1: Apply square - root property

We know that for non - negative real numbers \(m\) and \(n\), \(\sqrt{mn}=\sqrt{m}\cdot\sqrt{n}\), and \(\sqrt{m^{2}}=\vert m\vert\) (when \(m\) is a real number). Also, if \(m\) is non - negative, \(\sqrt{m^{2}} = m\). Assuming \(a\) and \(b\) are non - negative (since we are dealing with square roots of squares in a simplified form, we can assume the variables are non - negative for the purpose of simplification), we can use the property \(\sqrt{xy}=\sqrt{x}\cdot\sqrt{y}\) for \(x = a^{2}b^{2}\) and \(y=x\).

So, \(\sqrt{a^{2}b^{2}x}=\sqrt{a^{2}b^{2}}\cdot\sqrt{x}\)

Step2: Simplify \(\sqrt{a^{2}b^{2}}\)

We know that \(\sqrt{a^{2}b^{2}}=\sqrt{(ab)^{2}}\) (by the property \(\sqrt{m^{2}n^{2}}=\sqrt{(mn)^{2}}\)). And since we assumed \(a\) and \(b\) are non - negative, \(\sqrt{(ab)^{2}}=ab\) (because for non - negative \(k\), \(\sqrt{k^{2}}=k\))

So, \(\sqrt{a^{2}b^{2}}\cdot\sqrt{x}=ab\sqrt{x}\)

Answer:

\(ab\sqrt{x}\) (assuming \(a\geq0\) and \(b\geq0\))