Question

vă rog frumos cele două exerciții cu tot cu explicație pas cu pas va rog frumos urgent ​

Answer 1

16.

La inmultire cu aceeasi baza puterile se aduna, iar la impartire puterile se scad.

Un numar la o putere negativa- se inverseaza numarul si puterea devine pozitiva

$5^{\frac{2}{3} }\cdot 5^3\cdot 5^{\frac{3}{2} }=5^{\frac{31}{6}}$

$2^{\frac{1}{3} }\cdot 2^2=2^{\frac{7}{3} }$

$2^2\cdot 2^{\frac{1}{2} }=2^{\frac{5}{2}}$

$\frac{2}{2^{\frac{1}{3} }} =2\cdot 2^{-\frac{1}{3} }=2^{\frac{2}{3} }$

17.

a)

$\sqrt[3]{48} =\sqrt[3]{2^4\cdot 3} =2\sqrt[3]{6}$

$\sqrt[3]{135} =3\sqrt[3]{5}$

$\sqrt[3]{384} =4\sqrt[3]{6}$

$\sqrt[3]{40} =2\sqrt[3]{5}$

Rezultat: $-2\sqrt[3]{6} +\sqrt[3]{5}$

b)

$(\frac{\sqrt[3]{9} }{\sqrt[3]{216} } -3\sqrt[3]{3} +4\frac{1}{\sqrt[3]{3} } ):2\frac{1}{\sqrt[3]{3} } =\\\\=(\frac{1}{2\sqrt[3]{3} } -3\sqrt[3]{3}+4\frac{1}{\sqrt[3]{3} } )\cdot\frac{\sqrt[3]{3} }{2} =\\\\=\frac{1-6\sqrt[3]{9}+8 }{2\sqrt[3]{3} } \cdot\frac{\sqrt[3]{3} }{2} =\\\\=\frac{6\sqrt[3]{9}+9}{4}$

Un alt exercitiu gasesti aici: https://brainly.ro/tema/10035006

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