Question

решить домашнее задание 9 клас

Answer 1

1 задание

${300}^{ - 2} = \frac{1}{ {300}^{2} } = \frac{1}{90000}$

${3000}^{2} = 9000000$

${(0.002)}^{ - 3} = \frac{1}{ {(0.002)}^{3} } = \frac{1}{0.0000008}$

${(0.00098)}^{ - 2} = \frac{1}{ {(0.00098)}^{2} } = \frac{1}{0.0000009604}$

2 задание (а)

$(x + 2)(x - 3) \leqslant 0$

$(x + 2)(x - 3) = 0$

$x + 2 = 0, x - 3 = 0$

$x_1 = - 2, x_2 = 3$

$- 2 \geqslant x \geqslant 3$

[-2; 3]

2 задание (б)

$(x + 2) { (x - 1)}^{2} 0$

$(x + 2){ (x - 1)}^{2} = 0$

$x + 2 = 0, { (x - 1)}^{2} = 0$

$x = -2, x - 1 = 0$

$x = -2, x = 1$

$-2 < x < 1; x 1$

(-2; 1)U(1; +бесконечность)

2 задание (в)

$\frac{3 - x}{2 + x} \geqslant 0$

ОДЗ: 2+x≠0, x≠-2

$3 - x \geqslant 0$

$x \leqslant 3$

$x \leqslant 3, x ≠ -2$

(-бесконечность; -2)U(-2; 3]

2 задание (г)

$\frac{(x - 1)(x + 2)}{x} < 0$

ОДЗ: x ≠ 0

$(x - 1)(x + 2) = 0$

$x - 1 = 0, x + 2 = 0$

$x = 1, x = - 2$

-2 < x < 1, x ≠0

(-2; 0)U(0; 1)

3 задание (а)

$\sqrt{3} - 5 \sqrt{108} + \frac{1}{2} \sqrt{12} = \sqrt{3} - 5 \sqrt{3 \times 36} + \frac{1}{2} \sqrt{3 \times 4} = \sqrt{3} - 5 \times \sqrt{3} \times \sqrt{36} + \frac{1}{2} \times \sqrt{3} \times \sqrt{4} = \sqrt{3} - 5 \times \sqrt{3} \times 6 + \frac{1}{2} \times \sqrt{3} \times 2 = \sqrt{3} - 30 \sqrt{3} + \sqrt{3} = \sqrt{3} (1 - 30 + 1) = - 28\sqrt{3}$

3 задание (б)

$2 \sqrt{18} + 3 \sqrt{8} + 3 \sqrt{32} - \sqrt{50} = 2 \sqrt{2 \times 9} + 3 \sqrt{2 \times 4} + 3 \sqrt{2 \times 16} - \sqrt{2 \times 25} = 2 \times \sqrt{2} \times \sqrt{9} + 3 \times \sqrt{2} \times \sqrt{4} + 3 \times \sqrt{2} \times \sqrt{16} - \sqrt{2} \times \sqrt{25} = 2 \times \sqrt{2} \times 3 + 3 \times \sqrt{2} \times 2 + 3 \times \sqrt{2} \times 4 - \sqrt{2} \times 5 = 6 \sqrt{2} + 6 \sqrt{2} + 12 \sqrt{2} - 5 \sqrt{2} = \sqrt{2} (6 + 6 + 12 - 5) = 19 \sqrt{2}$

4 задание

$y = 3 + 2x - {x}^{2}$

$y = - {x}^{2} + 2x + 3$

$- {x}^{2} + 2x + 3 = 0$

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

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 2 + \sqrt{16} }{2 \times ( - 1)} = \frac{ - 2 + 4}{ - 2} = \frac{2}{ - 2} = - 1$

$x_1 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 2 - \sqrt{16} }{2 \times ( - 1)} = \frac{ - 2 - 4}{ - 2} = \frac{ - 6}{ - 2} = 3$

ответ: -1; 3