Question

Решите неравенства 1. √(5-2x)< 6x-1 2. log 3-2x x^2< =1 3.3√x-√(5x+5)> 1

Answer 1

$1)\; \; \sqrt{5-2x} \ \textless \ 6x-1\; ,\; \; \; ODZ:5-2x \geq 0\; ,\; x \leq 2,5\\\\5-2x\ \textless \ 36x^2-12x+1\\\\36x^2-10x-4\ \textgreater \ 0\\\\18x^2-5x-1\ \textgreater \ 0\\\\D=169\; ,\; \; x_1=-\frac{2}{9}\; ,\; \; x_2= \frac{1}{2} \\\\+++(-\frac{2}{9})---(\frac{1}{2})+++\quad x\in (-\infty ,- \frac{2}{9} )\cup ( \frac{1}{2},+\infty )\\\\x \leq 2,5\; \; \; \Rightarrow \; \; \; \underline {x\in (-\infty ,-\frac{2}{9})\cup( \frac{1}{2};\; 2,5] }$

$2)\; \; log_{3-2x}x^2\leq 1\; ,\; \; \; ODZ:\; \; \left \{ {{3-2x\ \textgreater \ 0} \atop {3-2x\ne 1}} \right. \; ,\left \{ {{x\ \textless \ 1,5} \atop {x\ne 1}} \right.$

Метод рационализации:  $log_{h}f$

$(3-2x-1)(x^2-3+2x) \leq 0\\\\-2(x-1)(x-1)(x+3) \leq 0\; |\cdot (-1/2)\\\\(x-1)^2(x+3) \geq 0\; \; \; ---[-3\; ]+++(1)+++(1,5)\\\\\underline {x\in [-3;1)\cdot (1;\; 1,5)}$

$3)\quad 3\sqrt{x}-\sqrt{5x+5}\ \textgreater \ 1\; ,\; \; \; ODZ:\; x \geq 0\\\\3\sqrt{x}\ \textgreater \ 1+\sqrt{5x+5}\\\\9x\ \textgreater \ 1+2\sqrt{5x+5}+5x+5\\\\2\sqrt{5x+5}\ \textless \ 4x-6\; |:2\\\\\sqrt{5x+5}\ \textless \ 2x-3\\\\5x+5\ \textless \ 4x^2-12x+9\\\\4x^2-17x+4\ \textgreater \ 0\\\\D=225\; ,\; \; x_1=\frac{1}{4}\; ,\; \; x_2=4\\\\\quad [\; 0\; ]+++(\frac{1}{4})---(4)+++\\\\\underline {x\in [\; 0,\frac{1}{4})\cup (4,+\infty )}$