Question

dividing and square roots
simplify.

1. \\(\frac{\sqrt{3}}{\sqrt{48}}\\)
2. \\(\frac{\sqrt{12}}{\sqrt{4}}\\)

1. \\(\frac{\sqrt{20}}{\sqrt{5}}\\)
2. \\(\frac{\sqrt{8}}{\sqrt{100}}\\)

1. \\(\frac{\sqrt{15}}{\sqrt{125}}\\)
2. \\(\frac{\sqrt{6}}{\sqrt{8}}\\)

1. \\(\frac{4\sqrt{6}}{3\sqrt{8}}\\)
2. \\(\frac{2\sqrt{3}}{4\sqrt{27}}\\)

1. \\(\frac{2\sqrt{3}}{2\sqrt{12}}\\)
2. \\(\frac{4\sqrt{6}}{4\sqrt{27}}\\)

1. \\(\frac{3\sqrt{20}}{4\sqrt{16}}\\)
2. \\(\frac{3\sqrt{20}}{3\sqrt{36}}\\)

Explanation:

Problem 1: $\boldsymbol{\frac{\sqrt{3}}{\sqrt{48}}}$

Step1: Simplify the denominator

We know that $\sqrt{48} = \sqrt{16\times3} = \sqrt{16}\times\sqrt{3} = 4\sqrt{3}$
So the expression becomes $\frac{\sqrt{3}}{4\sqrt{3}}$

Step2: Cancel out common terms

The $\sqrt{3}$ in the numerator and denominator cancels out, leaving $\frac{1}{4}$

Step1: Simplify numerator and denominator

$\sqrt{12} = \sqrt{4\times3} = 2\sqrt{3}$ and $\sqrt{4} = 2$
The expression becomes $\frac{2\sqrt{3}}{2}$

Step2: Cancel out common terms

The 2 in the numerator and denominator cancels out, leaving $\sqrt{3}$

Step1: Use the property of square roots $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$

So $\frac{\sqrt{20}}{\sqrt{5}}=\sqrt{\frac{20}{5}}$

Step2: Simplify the fraction inside the square root

$\frac{20}{5} = 4$, so we have $\sqrt{4}$

Step3: Evaluate the square root

$\sqrt{4}=2$

Answer:

$\frac{1}{4}$

Problem 2: $\boldsymbol{\frac{\sqrt{12}}{\sqrt{4}}}$