Respuesta :

[tex]\bf \cfrac{3^8a^{10}b^{-5}c^2}{3^{12}a^7b^{-3}c^{-2}}\implies \cfrac{a^{10}a^{-7}c^2c^2}{3^{12}\cdot 3^{-8}b^{-3}b^5}\implies \cfrac{a^{10-7}c^{2+2}}{3^{12-8}b^{-3+5}}\implies \cfrac{a^3c^4}{3^4b^2}~\hfill \begin{cases} a=4\\ b=8\\ c=3 \end{cases}[/tex]

[tex]\bf \cfrac{4^3\cdot ~~\begin{matrix} 3^4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 3^4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\cdot 8^2}\implies \cfrac{~~\begin{matrix} 64 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 64 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies 1[/tex]

gmany

Answer:

1

Step-by-step explanation:

[tex]\dfrac{3^8a^{10}b^{-5}c^2}{3^{12}a^7b^{-3}c^{-2}}\qquad\text{use}\ \dfrac{x^n}{x^m}=x^{n-m}\\\\=3^{8-12}a^{10-7}b^{-5-(-3)}c^{2-(-2)}=3^{-4}a^3b^{-2}c^{4}\\\\\text{substitute:}\ a=4,\ b=8,\ c=3:\\\\(3^{-4})(4^3)(8^{-2})(3^4)=(3^{-4}\cdot3^4)\bigg(2^2\bigg)^3\bigg(2^3\bigg)^{-2}\\\\\text{use}\ (x^n)(x^m)=x^{n+m}\ \text{and}\ \bigg(x^n\bigg)^m=x^{nm}\\\\=(3^{-4+4})\bigg(2^{(2)(3)}\bigg)\bigg(2^{(3)(-2)}\bigg)=(3^{0})(2^6)(2^{-6})\\\\\text{use}\ x^{-n}=\dfrac{1}{x^n}\ \text{and}\ (x^n)(x^m)=x^{n+m}[/tex]

[tex]=(1)\left(2^{6+(-6)}\right)=2^0=1\\\\\text{Used}\ a^0=1\ \text{for any real value of}\ a,\ \text{except 0}.[/tex]

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