Question

ПОМОГИТЕ ПОЖАЛУЙСТА, СРОЧНО!​

Answer 1

$\displaystyle\bf\left \{ {{19x-2x^{2} -42>0} \atop {(x^{2} -9)(x-4)(1-x)\leq 0}} \right.\\\\\\ \left \{ {{2x^{2} -19x+42<0} \atop {(x-3)(x+3)(x-4)(x-1)\geq 0}} \right. \\\\\\ 1) \ 2x^{2} -19x+42<0\\\\ 2x^{2} -19x+42=0\\\\ D=(-19)^{2} -4\cdot2\cdot42=361-336=25=5^{2} \\\\\\ x_{1} =\frac{19-5}{4} =3,5\\\\\\ x_{2} =\frac{19+5}{4} =6\\\\\\ 2x^{2} -19x+42=2(x-3,5)(x-6)\\\\ (x-3,5)(x-6)<0$

+ + + + + (3,5) - - - - - (6) + + + + +

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$\displaystyle\bf x\in(3,5 \ , \ 6)\\\\\\ 2) \ (x-3)(x+3)(x-4)(x-1)\geq 0$

+ + + + + [- 3] - - - - - [1] + + + + + [3] - - - - - [4] + + + + +

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$\displaystyle\bf x\in(-\infty \ ; -3] \ \cup \ [1 \ , \ 3] \ \cup \ [4 \ ; \ +\infty)\\\\\\ Otvet:x\in[4 \ ; \ 6)$