Question

es sobre teoria de exponentesssssss :)

Answer 1

Respuesta:

Recuerda :

=> $x^{m} .x^{n} =x^{m+n}$

=> $(x)^{-m}=\frac{1}{x^{m} }$

=> $(x^{m} )^{n}=x^{m.n}$ $\neq x^{m^{n} }$

=> $\frac{x^{m} }{x^{n} }=x^{m-n}$

Ahora :

1 . Reduce :

$\frac{x^{n-2}.x^{2n+1} .x^{2n+1} }{x^{n+2} .x^{3n-3} } =\frac{x^{n-2+2n+1+2n+1} }{x^{n+2+3n-3} } =\frac{x^{5n} }{x^{4n-1} } =x^{5n-(4n-1)}=x^{5n-4n+1}=x^{n+1}$

2. Calcula :

$P = \sqrt{(\frac{7}{4})^{-1} .28+(\frac{1}{4})^{-1} +(\frac{1}{\sqrt{5} })^{-2} } =...\\\\...=\sqrt{\frac{4}{7}.28+\frac{4}{1}+\frac{\sqrt{5}^{2} }{1^{2} } } =\sqrt{16+4+5 }=\sqrt{25}=5$

3. Reduce :

$R=\frac{(x^{2})^{5}.x^{3^{2} } .x^{2^{3} } }{x^{2^{5} }.x^{-6} } =\frac{x^{10}.x^{9 } .x^{8 } }{x^{32 }.x^{-6} }=\frac{x^{10+9+8} }{x^{32-6} } =\frac{x^{27} }{x^{26} } =x^{27-26}=x$

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Recuerda : Porque la Respuesta anterior esta mal :

$(x^{2} )^{3} =x^{2.3} =x^{6} \\x^{2^{3} } =x^{2.2.2} =x^{8} \\Entonces :\\\\(x^{2} )^{3} \neq x^{2^{3} }$