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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