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Question 1

Question 2

Решите задачу:

$1)\quad log_2^2x-4log_2x-1=0\; ,\; \; \; \; ODZ;\; x\ \textgreater \ 0\\\\t=log_2x\; ,\; \; \; t^2-4t-1=0\\\\D/4=4+1=5\; ,\; \; t_1=2-\sqrt5\; ,\; \; t_2=2+\sqrt5\\\\a)\; \; \; log_2x=2-\sqrt5\; ,\; \; x=2^{2-\sqrt5}\\\\b)\; \; \; log_2x=2+\sqrt5\; ,\; \; x=2^{2+\sqrt5}\\\\c)\; \; x_1\cdot x_2=2^{2-\sqrt5}\cdot 2^{2+\sqrt5}=2^{2-\sqrt5+2+\sqrt5}=2^4=16$

$2)\quad (cos\frac{\pi}{6})^{x^2-3x-10}=(sin\frac{\pi}{6})^{x^2-3x-10}\\\\(\frac{\sqrt3}{2})^{x^2-3x-10}\ \textless \ (\frac{1}{2})^{x^2-3x-10}\; |:(\frac{1}{2})^{x^2-3x-10}\\\\(\sqrt3)^{x^2-3x-10}\ \textless \ 1\\\\(\sqrt3})^{x^2-3x-10}\ \textless \ (\sqrt3})^0\\\\\sqrt3\ \textgreater \ 1\quad \Rightarrow \; \; \; x^2-3x-10\ \textless \ 0\; ,\; \; \; \; x_1=-2\; ,\; \; x_2=5\\\\(x-5)(x+2)\ \textless \ 0+++(-2)---(5)+++\\\\x\in (-2,5)$

$3)\; \; \; lg(x^2+2x-3)=lg(x-3)\; ,\\\\ODZ:\; \; \left \{ {{x^2+2x-3\ \textgreater \ 0} \atop {x-3\ \textgreater \ 0}} \right. \; \left \{ {{(x-1)(x+3)\ \textgreater \ 0} \atop {x\ \textgreater \ 3}} \right. \; \left \{ {{x\in (-\infty ,-3)\cup (1,+\infty )} \atop {x\in (3,+\infty )}} \right. \; \; \to \\\\x\in (3,+\infty )\\\\\\x^2+2x-3=x-3\\\\x^2+x=0\\\\x(x+1)=0\\\\x_1=0\notin ODZ\; ,\\\\ x_2=-1\notin ODZ\\\\Otvet:\; \; x\in \varnothing \; .$

Question 3

Спасибо!