Jawabanmu

2014-08-16T17:57:51+07:00
H. (\frac{24a^{3}.b^{8}}{6a^{5}.b}).(\frac{4b^{3}.a}{2a^{3}})^{2} \\ = \frac{4b^{7}}{a^{2}}.\frac{4b^{6}}{a^{4}} \\ = \frac{16b^{13}}{a^{6}}

j. (\frac{(-p)^{3}.(-q)^{2}.r^{3}}{-3(p^{2}q)^{2}}) : (\frac{2pqr^{3}}{-12(qr)^{2}}) \\ = \frac{-p^{3}q^{2}r^{3}}{-3p^{6}q^{3}} : \frac{pqr^{3}}{-6q^{2}r^{2}} \\ = \frac{r^{2}}{3p^{3}q} . (-\frac{6q}{r}) \\ = -\frac{2r^{2}}{p^{3}}
terimakasih :)
oh iya, itu yg (4b^3. a / 2a^3)^2 kok bisa hasilnya 4b^6/a^4?
dibagi dulu trs baru dikuadratin