Well, $$fracsqrt33 = fracsqrt3sqrt3sqrt3 =fracsqrt3sqrt3frac1sqrt3 =frac1sqrt3.$$

It"s really simply the fact that $$fracaa^2 =frac1a.$$



Let"s take into consideration a general case, for part integer $x$. (In fact, it"s true for any type of real number $x$.) us have$$ x^1/2 / x = x^1/2 cdot x^-1 = x^-1/2. $$


All the algebra over is correct, despite if friend haven"t learned exponent rules yet that could not it is in intuitive. Here"s how I think around it; think about the portion $dfracx^ax^b$. You have actually $a$ $x$"s on the top and also you have actually $b$ $x$"s ~ above the bottom. Division means "taking away", so us take away $b$ $x$"s from the top, which used to have actually $a$ the them. So now the top has actually $a-b$ $x$"s and the bottom is now 1, so

$$dfracx^ax^b = x^a-b$$

Notice just how I never ever actually stated $a$ and also $b$ were entirety numbers. This logic still applies if they space fractions or any type of real number. Therefore $dfracsqrtxx = x^1/2 - 1 = x^-1/2 = dfrac1sqrtx$.

You are watching: Square root of x divided by x

$$frac sqrt xx=frac 1sqrt x$$

since $$frac ab=frac cd iff ad=bc$$

with $bd eq0$

In general

$$frac sqrt x^bx=frac 1sqrt x^a-b$$

$($for $x>0)$

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