Question

допоможіть , будь ласка​

Answer 1

1)

$3 - 5x \leqslant 0$

$- 5x \leqslant - 3 \: \: | \times ( - 1)$

$5x \geqslant 3$

$x \geqslant \frac{3}{5}$

$x \in[ \frac{3}{5} ; + \infty )$

2)

$3 - 2x < 3(4x - 5)$

$3 - 2x < 12x - 15$

$- 2x - 12x < - 15 - 3$

$- 14x < - 18 \: \: | \times ( - 1)$

$14x > 18$

$x > \frac{18}{14}$

$x > \frac{9}{7}$

$x \in( \frac{9}{7} ; + \infty )$

3)

${(x - 2)}^{2} \leqslant (x - 3)(x + 3) - 10$

${x}^{2} - 4x + 2 \leqslant {x}^{2} - 9 - 10$

${x}^{2} - {x}^{2} - 4x \leqslant - 9 - 10 - 2$

$- 4x \leqslant - 21 \: \: | \times ( - 1)$

$4x \geqslant 21$

$x \geqslant \frac{21}{4}$

$x \in[ \frac{21}{4} ; + \infty )$

4)

${x}^{2} + x - 30 < 0$

${x}^{2} + x - 30 = 0$

$D = {b}^{2} - 4ac = {1}^{2} - 4 \times 1 \times ( - 30) = 1 + 120 = 121 = {11}^{2}$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 1 + 11}{2 \times 1} = \frac{10}{2} = 5$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 1 - 11}{2 \times 1} = \frac{ - 12}{2} = - 6$

<приложение1>

$x \in( - 6;5)$

5)

$(x - 4)(x + 4)(x + 8) \geqslant 0$

$(x - 4)(x + 4)(x + 8) = 0$

$x - 4 = 0,x + 4 = 0,x + 8 = 0$

$x = 4, \: x = - 4, \: x = - 8$

<приложение2>

$x \in( - \infty ; + \infty )$

6)

$( {x}^{2} + 2x - 3)(x + 7) < 0$

$( {x}^{2} + 2x - 3)(x + 7) = 0$

${x}^{2} + 2x - 3 = 0, \: x + 7 = 0$

$D = {b}^{2} - 4ac = {2}^{2} - 4 \times 1 \times ( - 3) = 4 + 12 = 16 = {4}^{2}$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 2 + 4}{2 \times 1} = \frac{2}{2} = 1$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 2 - 4}{2 \times 1} = \frac{ - 6}{2} = - 3$

$x + 7 = 0$

$x_3 = - 7$

<приложение3>

$x \in( - \infty ; - 7)U( - 3;1)$

7)

$(2x + 7)(x - 2) - (x + 4)(x - 5) \leqslant 18$

$2 {x}^{2} - 4x + 7x - 14 - {x}^{2} + 5x - 4x + 20 \leqslant 18$

${x}^{2} + 4x - 12 \leqslant 0$

${x}^{2} + 4x - 12 = 0$

$D = {b}^{2} - 4ac = {4}^{2} - 4 \times 1 \times ( - 12) = 16 + 48 = 64 = {8}^{2}$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 4 + 8}{2 \times 1} = \frac{4}{2} = 2$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 4 - 8}{2 \times 1} = \frac{ - 12}{2} = - 6$

<приложение4>

$x \in[2; + \infty )$