Školski Forum

Riješiti jednačinu :
$$ \Large {
\sqrt{2} \cdot 0.5^{\frac{5}{4 \cdot \sqrt{x} +10}}=4^{\frac{1}{\sqrt{x}+1}}
}$$

bostan je napisao/la: ↑29 vel 2024, 18:52 Riješiti jednačinu :
$$ \Large {
\sqrt{2} \cdot 0.5^{\frac{5}{4 \cdot \sqrt{x} +10}}=4^{\frac{1}{\sqrt{x}+1}}
}$$

Može da se napiše kao :
$$ \Large {
4^{\frac{1}{4}} \cdot 4^{-\frac{1}{2} \cdot \frac{5}{4 \cdot \sqrt{x} +10}}=4^{\frac{1}{\sqrt{x}+1}}
}$$
pa slijedi
$$ \Large {
4^{\frac{1}{4}-\frac{1}{2} \cdot \frac{5}{4 \cdot \sqrt{x} +10}}=4^{\frac{1}{\sqrt{x}+1}}
}$$
i
$$ \Large {
\frac{1}{4}-\frac{1}{2} \cdot \frac{5}{4 \cdot \sqrt{x} +10}=\frac{1}{\sqrt{x}+1}
}$$

Izdvajanjem ...
$$ \Large {
\frac{1}{4}-\frac{1}{4} \cdot \frac{5}{2 \cdot \sqrt{x} +5}=\frac{1}{\sqrt{x}+1}
}$$

dobijamo
$$ \Large {
\frac{1}{4} \left(1-\frac{5}{2 \cdot \sqrt{x} +5} \right)=\frac{1}{\sqrt{x}+1}
}$$
i
$$ \Large {
1-\frac{5}{2 \cdot \sqrt{x} +5}=\frac{4}{\sqrt{x}+1}
}$$

$$ \Large {
\frac{2 \cdot \sqrt{x} +\cancel{5}-\cancel{5}}{2 \cdot \sqrt{x} +5}=\frac{4}{\sqrt{x}+1}
}$$

$$ \Large {
\frac{2 \cdot \sqrt{x}}{2 \cdot \sqrt{x} +5}=\frac{4}{\sqrt{x}+1}
}$$

$$ \Large {
2 \cdot \sqrt{x} (\sqrt{x}+1) {}=4 (2 \cdot \sqrt{x} +5)
}$$

$$ \Large {
\sqrt{x} (\sqrt{x}+1) =2 (2 \cdot \sqrt{x} +5)
}$$

$$ \Large {
x+\sqrt{x}+1=4 \sqrt{x} +10
}$$

$$ \Large {
x-3\sqrt{x}-9=0
}$$