Question

ДАМ 100 БАЛЛОВ ,Помогите решить все задачи пожалуйста

Answer 1

3.

$\displaystyle\bf\\\left \{ {{3 {x}^{2} + 4x + 1 \geqslant 0 } \atop {5x - 1 < 0 }} \right. \\ \\ 1) \: 3 {x}^{2} + 4x + 1 = 0 \\ 3 {x}^{2} + 4x + 1 = 0 \\ a = 3\\ b =4 \\ c = 1 \\ D = {b}^{2} - 4ac = {4}^{2} - 4 \times 3 \times 1 = 16 - 12 = 4 \\ x_{1} = \frac{ - 4 - 2}{2 \times 3} = - \frac{6}{6} = - 1\\ x_{2} = \frac{ - 4 + 2}{2 \times 3} = - \frac{2}{6} = - \frac{1}{3} \\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ 3 {x}^{2} +4 x + 1 =3 (x + 1)(x + \frac{1}{3} ) \\ \\ (x + 1)(x + \frac{1}{3} ) \geqslant 0 \\ + + + [ - 1] - - - [ - \frac{1}{3} ] + + + \\ x \leqslant - 1 \: \: \: and \: \: \: x \geqslant - \frac{1}{3} \\ \\ 2) \: 5x - 1 < 0 \\ 5x < 1 \\ x < 0.2 \\ \displaystyle\bf\\3) \: \left \{ {{x \leqslant - 1 \: \: \: and \: \: \: x \geqslant - \frac{1}{3} } \atop {x < 0.2 }} \right. \\ \\ x \: \epsilon \: ( - \propto; \: - 1]U[ - \frac{1}{3} ; \: 0.2)$

4.

$\frac{(2 - x) {}^{2} }{2 {x}^{2} - 5x + 2 } \\ 2 {x}^{2} - 5x + 2 = 0 \\ a = 2\\ b = - 5 \\ c = 2\\ D = {b}^{2} - 4ac = ( - 5) {}^{2} - 4 \times 2 \times 2 = 25 - 16 = 9 \\ x_{1} = \frac{5 - 3}{2 \times 2} = \frac{2}{4} = 0.5\\ x_{2} = \frac{5 + 3}{2 \times 2} = \frac{8}{4} = 2 \\ 2 {x}^{2} - 5x + 2 = 2(x - 0.5)(x - 2) \\ \\ \frac{(x - 2) {}^{2} }{(x - 0.5)(x - 2)} \\ (x - 2) {}^{3} (x - 0.5)$

Нет знака неравенства

5.

$\displaystyle\bf\\\left \{ {{2 {x}^{2} - 5x + 3 \geqslant 0 } \atop { {x}^{2} < 64 }} \right. \\ \\ 1) \: 2 {x}^{2} - 5x + 3 \geqslant 0 \\ 2 {x}^{2} - 5x + 3 = 0 \\ a = 2 \\ b = - 5 \\ c = 3\\ D = {b}^{2} - 4ac = ( - 5) {}^{2} - 4 \times 2 \times 3 = 25 - 24 = 1 \\ x_{1} = \frac{5 - 1}{2 \times 2} = \frac{4}{4} = 1 \\ x_{2} = \frac{5 + 1}{2 \times 2} = \frac{6}{4} = \frac{3}{2} = 1.5 \\ 2 {x}^{2} - 5x + 3 = 2(x - 1)(x - 1.5) \\ \\ (x - 1)(x - 1.5) \geqslant 0 \\ + + + [1] - - - [1.5] + + + \\ x \leqslant 1 \: \: \: and \: \: \: x \geqslant 1.5 \\ \\ 2) \: {x}^{2} < 64 \\ {x}^{2} - 64 < 0 \\ (x - 8)(x + 8) < 0 \\ + + + ( - 8) - - - (8) + + + \\ - 8 < x < 8 \\ \displaystyle\bf\\3) \: \left \{ {{x \leqslant 1 \: \: \: and \: \: \: x \geqslant 1.5} \atop { - 8 < x < 8 }} \right. \\ \\ x \: \epsilon \: ( - 8; \: 1]U[1.5; \: 8)$