Question

ПОМОГИТЕ ПОЖАЛУЙСТА,нужно срочно!!3 задание даю много баллов

Answer 1

${x}^{2} - 4x + 3 = 0 \\ \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1} + x_{2} = 4\\ x_{1} x_{2} =3 \\ x_{1} = 1\\ x_{2} = 3 \\ {x}^{2} - 4x + 3 = (x - 1)(x - 3)$

$\displaystyle\bf\\\left \{ {{ {x}^{2} - 4x + 3 \leqslant 0 } \atop {2x - 4 < 0 }} \right. \\ \displaystyle\bf\\\left \{ {{(x - 1)(x - 3) \leqslant 0} \atop {x < 2 }} \right. \\ \\ (x - 1)(x - 3) \leqslant 0 \\ + + + [1] - - - [3] + + + \\ 1 \leqslant x \leqslant 3 \\ \displaystyle\bf\\\left \{ {{1 \leqslant x \leqslant 3} \atop {x < 2 }} \right. \\ \\ otvet \: \: \: x \:\epsilon \: [1; \: 2)$