Задание 2.16

Question

2.16. Решите уравнение:
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1) $8^{\frac{2}{x}}-2^{\frac{2 x+3}{x}}-32=0\qquad\qquad $
3) $2^{\cos 2 x}-3 \cdot 2^{\cos ^{2} x}+4=0$.
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2) $5^{\sqrt{x-2}}-5^{1-\sqrt{x-2}}-4=0$;

Gimli · Accepted Answer

\begin{enumerate}
\item $$8^{\frac{2}{x}}-2^{\frac{2 x+3}{x}}-32=0 $$
$$2^{\frac{6}{x}}-2^{\frac{2 x+3}{x}}-32=0 $$
$$2^{\frac{6}{x}}-2^{2+\frac{3}{x}}-2^{5}=0 $$
$$2^{\frac{6}{x}}-2^{2} \cdot 2^{\frac{3}{x}}-32=0 $$
$$\left(2^{\frac{3}{2}}ight)^{2}-4 \cdot 2^{\frac{3}{x}}-32=0$$
Сделаем замену $2^{\frac{3}{x}}=t$:
$$t^{2}-4 t-32=0 $$
$$D=16+128=144=12^{2} $$
$$t_{1}=\frac{4+12}{2}=8, t_{2}=\frac{4-12}{2}=-4$$
गा.к. никогда не является правдой  высказывание $2^{\frac{3}{x}}=-4$, то
$$2^{\frac{3}{x}}=8 \Leftrightarrow \frac{3}{x}=3 \Leftrightarrow x=1 $$
	extbf{Ответ:  $x=\{1\}$}

\item $$5^{\sqrt{x-2}}-5^{1-\sqrt{x-2}}-4=0$$
$$5^{\sqrt{x-2}}-5^{1}: 5^{\sqrt{x-2}}-4=0 $$
$$5^{\sqrt{x-2}}-\frac{5}{5^{\sqrt{x-2}}}-4=0$$
Сделаем замену $5^{\sqrt{x-2}}=t$:
$$t-\frac{5}{t}-4=0$$
$$t^{2}-5-4 t=0 $$
$$t^{2}-4 t-5=0 $$
$$D=16+20=36=6^{2} $$
$$t_{1}=\frac{4+6}{2}=5, \quad t_{2}=\frac{4-6}{2}=-1$$

गा.к. $5^{\sqrt{x-2}}=-1$, то
$$5^{\sqrt{x-2}}=5 $$
$$\sqrt{x-2}=1 $$
$$x=\{3\}$$
	extbf{Ответ: $x=\{3\}$.}
\item $$2^{\cos 2 x}-3 \cdot 2^{\cos ^{2} x}+4=0$$ 
$$2^{\cos 2 x}-3 \cdot 2^{\frac{1+\cos 2x}{2}}+4=0$$
$$2^{\cos 2 x}-3 \cdot 2^{\frac{1}{2}} \cdot 2^{\cos 2 x}+4=0$$
$$2^{\cos 2 x}-3 \sqrt{2} \cdot 2^{\frac{\cos 2 x}{2}}+4=0$$
Сделаем замену $2^{\cos ^{2} x}$ на $t$:
$$t^2 - 3\sqrt{2}t + 4 = 0$$
$$D = b^2 - 4ac = 18 - 16 = 2$$
$$t_1 = \frac{3\sqrt{2} + \sqrt{2}}{2} = 2\sqrt{2}, \quad t_2 = \sqrt{2}.$$
Решим полученную совокупность:
$$\left[\begin{array}{rc1}{t_1 = 2\sqrt{2}}\{t_2 = \sqrt{2}}\end{array}ight.\Leftrightarrow \left[\begin{array}{rc1}{2^{\frac{\cos 2x}{2}} = 2\sqrt{2}}\{2^{\frac{\cos 2x}{2}} = \sqrt{2}}\end{array}ight.\Leftrightarrow \left[\begin{array}{rc1}{2^{\cos 2x} = 8}\{2^{\cos 2x} = 2}\end{array}ight.\Leftrightarrow \left[\begin{array}{rc1}{\cos 2x = 3}\{\cos 2x = 1}\end{array}ight.\Leftrightarrow \left[\begin{array}{rc1}{x = \varnothing}\{2x = 2\pi n}\end{array}ight. \Leftrightarrow \left[\begin{array}{rc1}{x = \varnothing}\{x = \pi n}\end{array}ight.$$
	extbf{Ответ: $x = \pi n, \quad n\in \mathbb{Z}$}
\end{enumerate}

Задание 2.16

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Задание 2.16

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