34. Poiščite tak $$ y $$, da bo $$ \frac{\sqrt[3]{a b^{-2}} \cdot \sqrt[6]{a^{-5} b}}{\left(\sqrt[12]{a^{-1} b^{3}}\right)^{2}}=\frac{1}{y \cdot \sqrt[3]{a}} $$

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$$
\textbf{Korak 1.} \text{ Zapišimo vse korenine z eksponentno obliko.}
$$

$$
\sqrt[3]{a b^{-2}} = (a b^{-2})^{\frac{1}{3}} = a^{\frac{1}{3}} b^{-\frac{2}{3}},
$$
$$
\sqrt[6]{a^{-5} b} = (a^{-5} b)^{\frac{1}{6}} = a^{-\frac{5}{6}} b^{\frac{1}{6}},
$$
$$
\sqrt[12]{a^{-1} b^{3}} = (a^{-1} b^{3})^{\frac{1}{12}} = a^{-\frac{1}{12}} b^{\frac{3}{12}} = a^{-\frac{1}{12}} b^{\frac{1}{4}}.
$$

$$
\textbf{Korak 2.} \text{ Združimo izraze v števcu:}
$$

$$
\sqrt[3]{a b^{-2}} \cdot \sqrt[6]{a^{-5} b} = a^{\frac{1}{3}}b^{-\frac{2}{3}} \cdot a^{-\frac{5}{6}}b^{\frac{1}{6}} = a^{\frac{1}{3} - \frac{5}{6}} b^{-\frac{2}{3}+ \frac{1}{6}}.
$$

$$
\frac{1}{3} - \frac{5}{6} = \frac{2}{6} - \frac{5}{6} = -\frac{3}{6} = -\frac{1}{2},
$$
$$
-\frac{2}{3} + \frac{1}{6} = -\frac{4}{6} + \frac{1}{6} = -\frac{3}{6} = -\frac{1}{2}.
$$

$$
\text{Torej v števcu dobimo: } a^{-\frac{1}{2}} b^{-\frac{1}{2}}.
$$

$$
\textbf{Korak 3.} \text{ Obdelamo imenovalec.}
$$

$$
\left(\sqrt[12]{a^{-1} b^{3}}\right)^{2} = \left(a^{-\frac{1}{12}}b^{\frac{1}{4}}\right)^{2} = a^{-\frac{1}{6}}b^{\frac{1}{2}}.
$$

$$
\textbf{Korak 4.} \text{ Izračunamo celoten levi člen ulomka:}
$$

$$
\frac{a^{-\frac{1}{2}}b^{-\frac{1}{2}}}{a^{-\frac{1}{6}}b^{\frac{1}{2}}} = a^{-\frac{1}{2} + \frac{1}{6}} b^{-\frac{1}{2} - \frac{1}{2}}.
$$

$$
-\frac{1}{2} + \frac{1}{6} = -\frac{3}{6} + \frac{1}{6} = -\frac{2}{6} = -\frac{1}{3},
$$
$$
-\frac{1}{2} - \frac{1}{2} = -1.
$$

$$
\text{Torej je izraz enak: } a^{-\frac{1}{3}} b^{-1}.
$$

$$
\textbf{Korak 5.} \text{ Postavimo enačbo glede na dano zvezo:}
$$

$$
\frac{a^{-\frac{1}{3}} b^{-1}}{1} = \frac{1}{y \cdot a^{\frac{1}{3}}}.
$$

$$
\textbf{Korak 6.} \text{ Pomnožimo obe strani enačbe z } y \cdot a^{\frac{1}{3}}:
$$

$$
y \cdot a^{\frac{1}{3}} \cdot a^{-\frac{1}{3}} b^{-1} = 1.
$$

$$
a^{\frac{1}{3}} \cdot a^{-\frac{1}{3}} = a^{0} = 1 \quad \Rightarrow \quad y \cdot b^{-1} = 1.
$$

$$
\textbf{Korak 7.} \text{ Rešimo za } y:
$$

$$
y = b.
$$

$$
\boxed{y = b.}
$$
$$ y = b $$

Poiščite tak , da bo

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