Question

решите систему неравенств ПОМОГИТЕ СРОЧНО 50б​

Answer 1

$\displaystyle\bf\\\left \{ {{5x-12 < 0} \atop {3x^{2} -7x+4\geq 0}} \right. \\\\\\1)\\\\5x-12 < 0\\\\5x < 12\\\\\boxed{x < 2,4}\\\\2)\\\\3x^{2} -7x+4\geq 0\\\\3x^{2} -7x+4=0\\\\D=(-7)^{2} -4\cdot 3\cdot 4=49-48=1\\\\\\x_{1} =\frac{7-1}{6} =1\\\\\\x_{2} =\frac{7+1}{6} =1\frac{1}{3} \\\\\\3x^{2} -7x+4=3\cdot\Big(x-1\Big)\cdot\Big(x-1\frac{1}{3} \Big)\\\\\\\Big(x-1\Big)\cdot\Big(x-1\frac{1}{3} \Big)\geq 0\\\\\\+ + + + + \Big[1\Big]- - - - - \Big[1\frac{1}{3} \Big]+ + + + +$

$\displaystyle\bf\\\boxed{x\in\Big(-\infty \ ; \ 1\Big]\cup\Big[1\frac{1}{3} \ ; \ +\infty\Big)}\\\\\\Otvet \ : \ x\in\Big(-\infty \ ; \ 1\Big]\cup\Big[1\frac{1}{3} \ ; \ 2,4\Big)$