Question

PORFA AYUDENME
DOY CORONA

Answer 1

Reduce M:

$(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}+(\frac{1}{8})^{-3}-(\frac{1}{5})^{-2}\\3^2+4^2+8^3-5^2\\9+16+512-25\\537-25\\512$

Aplica las propiedades y efectúa:

a) $(\frac{y^3}{a})^3(\frac{y^3}{a})^5=(\frac{y^3}{a})^8=\frac{y^{24}}{a^8}$

b) $(2.63)^9(2.63)^{11}(2.63)^3=(2.63)^{23}$

c) $[(\frac{c}{2d})^2]^3=(\frac{c}{2d})^6=\frac{c^6}{2d^6}$

d) $(\frac{5}{4})^9(\frac{2}{5})^3(\frac{2}{5})^6=(\frac{5}{4})^9(\frac{2}{5})^9=(\frac{5}{4}\cdot\frac{2}{5})^9=(\frac{10}{20})^9=(\frac{1}{2})^9$

e) $16^3\div2^3=(16\div2)^3=8^3=512$

f) $[(\sqrt28)^3]^0=\sqrt28^{3\cdot0}=\sqrt28^0=1$

g) $a^n\cdot b^n\cdot4^n=(a\cdot b\cdot4)^n$

h) $(\frac{x}{y})^3(\frac{x}{y})^9\div(\frac{x}{y})^2=(\frac{x}{y})^{3+9-2}=(\frac{x}{y})^{10}$

i) $[(2)^2]\div(2)^3[(2)^0]^9=(2)^{3+0\cdot9-2}=2^{3-2}=2^1=2$