Question

procedimiento plis
doy coronita si me ayudan​

Answer 1

Verified answer

Respuesta:

Explicación paso a paso:

[tex]\Large\underline{\underline{\textbf{Leyes de exponentes}}}[/tex]

► [tex]\texttt{Problema 1}[/tex]

1. [tex]\mathsf{(5^{3}.2^{2}.4^{3})^{2}=(5^{3}.2^{2}.(2^{2})^{3})^{2}=(5^{3}.2^{2}.2^{6})^{2}=(5^{3}.2^{8})^{2}=5^{6}.2^{16}}[/tex]
2. [tex]\mathsf{3^{3}.3^{2}\Big(\dfrac{3^{3}.3^{4}}{3^{4}.3^{2}}\Big)=3^{5}\Big(\dfrac{3^{7}}{3^{6}}\Big)=3^{5}(3^{1})=3^{6}}[/tex]
3. [tex]\mathsf{\Big(\dfrac{3.2^{3}}{2^{2}.3^{2}}\Big)^{2}=(3^{-1}.2^{1})^{2}=3^{-2}.2^{2}}[/tex]

► [tex]\texttt{Problema 2}[/tex]

1. [tex]\mathsf{(5^{2})^{3}=5^{6}}[/tex]
2. [tex]\mathsf{\bigg(\Big(\dfrac{3}{7}\Big)^{2}\bigg)^{5}=\Big(\dfrac{3}{7}\Big)^{10}}[/tex]
3. [tex]\mathsf{(\dfrac{4}{5})^{4}=\dfrac{256}{625}}[/tex]

► [tex]\texttt{Problema 3}[/tex]

1. [tex]\mathsf{\sqrt[3]{\dfrac{81}{3}}.3^{4}.\dfrac{1}{3^{4}}=\sqrt[3]{27}.\dfrac{3^{4}}{3^{4}}=3.1=3}[/tex]
2. [tex]\mathsf{\dfrac{(\sqrt{49}.\sqrt{49}.\sqrt{49}.\sqrt{49})^{2}}{((\sqrt{49})^{2})^{3}}=\dfrac{(7.7.7.7)^{2}}{((7)^{2})^{3}}=\dfrac{(7^{4})^{2}}{7^{6}}=\dfrac{7^{8}}{7^{6}}=7^{2}=49}[/tex]
3. [tex]\mathsf{\dfrac{10^{2}.10^{3}.\sqrt[3]{1000}.10^{4}}{(\sqrt[4]{10000}.\sqrt[3]{100})^{2}}=\dfrac{10^{2}.10^{3}.10.10^{4}}{(10.10)^{2}}=\dfrac{10^{10}}{(10^{2})^{2}}=\dfrac{10^{10}}{10^{4}}=10^{6}}[/tex]