Question

решить неравенство вида:
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Answer 1

$\frac{\left ( 5x+4 \right )\left ( 3x-2 \right )}{x+3}\leq \frac{\left ( 3x-2 \right )\left ( x+2 \right )}{1-x}\\\frac{\left ( 3x-2 \right )\left ( \left ( 1-x \right )\left ( 5x+4 \right ) \right )-\left ( x+3 \right )\left ( x+2 \right )}{\left ( x+3 \right )\left ( 1-x \right )}\leq 0\\\frac{\left ( 3x-2 \right )\left ( -6x^2-4x-2 \right )}{\left ( x+3 \right )\left ( 1-x \right )}\leq 0\Leftrightarrow \frac{\left ( 3x-2 \right )\left ( 3x^2+2x+1 \right )}{\left ( x+3 \right )\left ( 1-x \right )}\geq 0\\\frac{3x-2}{\left ( x+3 \right )\left ( 1-x \right )}\geq 0\Rightarrow x\in \left ( -\infty ;-3 \right )\cup \left [ \frac{2}{3};1 \right )$