I suppose you meant z > 0, or meant to use x's in place of z'sā¦
If z > 0, then
[tex]\dfrac{20\sqrt{z^6}}{\sqrt{16z^7}} = \dfrac{20\sqrt{\left(z^3\right)^2}}{\sqrt{4^2\left(z^3\right)^2z}} = \dfrac{20\left|z^3\right|}{4\left|z^3\right|\sqrt{z}} = \boxed{\dfrac{5}{\sqrt{z}}}[/tex]
because
[tex]\sqrt{z^2} = |z|[/tex]
and if z is positive, we have |z| = z, and we can eliminate common factors of z in the fraction,
[tex]\dfrac{\left|z^3\right|}{\left|z^3\right|} = \dfrac{z^3}{z^3} = 1[/tex]
