Question

4) 2 +4 a +3 2 + 1 a + 1 < 3 a + 2​

Answer 1

$\displaystyle\bf\\\frac{2}{a+3} +\frac{1}{a+1} < \frac{3}{a+2} \\\\\\\frac{2}{a+3} +\frac{1}{a+1}-\frac{3}{a+2} < 0\\\\\\\frac{2\cdot(a+1)(a+2)+1\cdot(a+3)(a+2)-3\cdot(a+3)(a+1)}{(a+3)(a+1)(a+2)} < 0\\\\\\\frac{2\cdot(a^{2} +3a+2)+a^{2} +5a+6-3\cdot(a^{2}+4a+3) }{(a+3)(a+2)(a+1)} < 0\\\\\\\frac{2a^{2} +6a+4+a^{2} +5a+6-3a^{2}-12a-9 }{(a+3)(a+2)(a+1)} < 0\\\\\\\frac{-a+1}{(a+3)(a+1)(a+2)} < 0\\\\\\\frac{a-1}{(a+3)(a+1)(a+2)} > 0$

Решим неравенство методом интервалов :

$\displaystyle\bf\\+ + + + + (-3)- - - - - (-2)+ + + + + (-1)- - - - - (1)+ + + + +$

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$\displaystyle\bf\\Otvet \ : \ a\in(-\infty \ ; \ -3) \ \cup (-2 \ ; \ -1) \ \cup \ (1 \ ; \ +\infty)$