 # To calculate: The simplified form of the expression displaystylefrac{sqrt{{12}}}{sqrt{{{x}+{1}}}}. Wribreeminsl 2020-12-16 Answered
To calculate: The simplified form of the expression $\frac{\sqrt{12}}{\sqrt{x+1}}.$
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Formula used:
Product property of radicals: $\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}$
Power property of radicals: ${\left({a}^{m}\right)}^{n}={a}^{mn}$
Calculation:
Consider the provided expression, $\frac{\sqrt{12}}{\sqrt{x+1}}$
Multiply numerator and denominator by $\sqrt{x+1}$ so that the radicand in the denominator is a perfect square.
Therefore,
$\frac{\sqrt{12}}{\sqrt{x+1}}=\frac{\sqrt{12}}{\sqrt{x+1}}\cdot \frac{\sqrt{x+1}}{\sqrt{x+1}}$
Here, $x>-1$
Apply the product property of radicals:
$\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}$
Therefore,
$\frac{\sqrt{12}}{\sqrt{x+1}}=\frac{\sqrt{4\cdot 3}\sqrt{x+1}}{\sqrt{\left(x+1\right)\left(x+1\right)}}$
$=\frac{\sqrt{3}\sqrt{x+1}}{\sqrt{{\left(x+1\right)}^{2}}}$
Apply power property
Therefore,
$\sqrt{{\left(x+1\right)}^{2}}={\left({\left(x+1\right)}^{2}\right)}^{\frac{1}{2}}$
$={\left(x+1\right)}^{2\cdot \frac{1}{2}}$
$={\left(x+1\right)}^{1}$
$=\left(x+1\right)$
Therefore,
$\frac{\sqrt{12}}{\sqrt{x+1}}=\frac{\sqrt{3}\sqrt{x+1}}{x+1}$
Apply the product property of radicals: $\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}$ in the numerator,
$\sqrt{3}\sqrt{x+1}=\sqrt{3\left(x+1\right)}$
$=\sqrt{3x+3}$
Hence,
$\frac{\sqrt{12}}{\sqrt{x+1}}=\frac{\sqrt{3x+3}}{x+1}$
Therefore, the simplified form of the expression

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