\displaystyle\sqrt{{\frac{{{2}{x}}}{{{9}{x}}}}}=\frac{\sqrt{{{2}}}}{{3}} excluding \displaystyle{x}={0} Explanation: For any positive Real numbers \displaystyle{a},{b} : \displaystyle\sqrt{{\frac{{a}}{{b}}}}=\frac{\sqrt{{{a}}}}{\sqrt{{{b}}}} ...

\displaystyle\frac{{{3}\sqrt{{{2}{x}}}}}{{x}} Explanation: \displaystyle\frac{{6}}{{\sqrt{{{2}{x}}}}} \displaystyle\frac{{6}}{{\sqrt{{{2}{x}}}}}\cdot{\left[\frac{{\cancel{-}\sqrt{{{2}{x}}}}}{{\cancel{{-}}\sqrt{{{2}{x}}}}}\right]} ...

\displaystyle{x}={2} Explanation: First, start by taking a look at the equation \displaystyle\sqrt{{{x}-{1}}}=\frac{{x}}{{2}} Right from the start, you need \displaystyle{x} to be ...

450 Explanation: \displaystyle\frac{\sqrt{{{2}{x}}}}{{5}}={6} \displaystyle\sqrt{{{2}{x}}}={30} \displaystyle{2}{x}={30}^{{2}} \displaystyle{2}{x}={900} \displaystyle{x}={450}

Hint: \begin{align*} \left|\frac{\sqrt{x}}{x-1}-\frac{\sqrt{x_0}}{x_0-1}\right|&= \left|\frac{\sqrt{x}}{x-1}+\frac{\sqrt{x}}{x_0-1}-\frac{\sqrt{x}}{x_0-1}-\frac{\sqrt{x_0}}{x_0-1}\right|\\ &\leq ...

\displaystyle{x}\in{\left(-\infty,{0}\right]}\cup{\left({2},+\infty\right)} Explanation: The first thing to look out for is any value of \displaystyle{x} that will maek the denominator of ...