Question

Момогите плз с 5 вариантом, по возможности с 6

Answer 1

Объяснение:

5 вариант

1. (7¹¹ · 7⁻¹²)² = (7⁻¹)² = 7⁻² = $\frac{1}{49}$.

2. $\frac{12\sqrt{45}}{\sqrt{5}} = \frac{12\sqrt{45}\sqrt{5}}{5} = \frac{12\sqrt{45*5}}{5} = \frac{12\sqrt{5*5*9}}{5} = \frac{12*5*3}{5} = 12*3 = 36$.

3. 5,8(1 - а) - 1,8(6 - а) < 5

5,8 - 5,8a - 10,8 + 1,8a < 5

-5,8a + 1,8a < 5 - 5,8 + 10,8

-4a < 10

a > -2,5

x∈ (-2,5; ∞)

4.1) 2x² - 5x - 3 = 0

D = 25 + 24 = 49

$x_1 = \frac{5-7}{4} = -\frac{1}{2}$

$x_2 = \frac{5+7}{4} = 3$

4.2) 3x² - 8x + 5 = 0

D = 64 - 60 = 4

$x_1 = \frac{8-2}{6} = 1$

$x_2 = \frac{8+2}{6} = \frac{10}{6} = \frac{5}{3} = 1\frac{2}{3}$

6 вариант.

1. (8⁹ · 8⁻⁸)⁻² = (8⁻¹)⁻² = 8² = 64.

2. $\frac{7\sqrt{8}}{\sqrt{98}} = \frac{7\sqrt{8}\sqrt{98}}{98} = \frac{7\sqrt{8*98}}{98} = \frac{7\sqrt{4*2*2*49}}{98} = \frac{7*4*7}{98} = 2$.

3. 8 - 4(2 - 5а) > - (5a + 6)

8 - 8 + 20a > - 5a + 6

20a + 5a > 6  

25a > 6

$a > \frac{6}{25} = 0,24$

x∈ (0,24; ∞)

4.1) 8x² - 14x + 5 = 0

D = 196 - 160 = 36

$x_1 = \frac{14-6}{16} = \frac{1}{2}$

$x_2 = \frac{14+16}{16} = \frac{15}{8} = 1\frac{7}{8}$

4.2) 12x² + 16x - 3 = 0

D = 256 + 144 = 400

$x_1 = \frac{-16-20}{24} = \frac{-9}{6} = -1\frac{1}{2}$

$x_2 = \frac{-16+20}{24} = \frac{4}{24} = \frac{1}{6}$