Question

Решить!
1)4 - 4x2 =0
2)16x2 + 22x = 38
3)x2 = 30 + x
4)16 - 8x + x2 =0
5)5x2 - 26x + 5 =0
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Answer 1

1) ± 1

2) 1; $-2\frac{3}{8}$

3) 6; - 5

4) 4

5) 5; 0, 2

Объяснение:

1) 4 - 4х² = 0

- 4х² = - 4

4х² = 4

х² = 1

х = ± 1

2) 16х² + 22х = 38

16х² + 22х - 38 = 0

8х² + 11х - 19 = 0

D = b² - 4ac = 121 + 608 = 729 = 27² ⇒ 2 корня:

$x_{1} = \frac{- b + \sqrt{D} }{2a} = \frac{ - 11 + 27}{16} = \frac{16}{16} = 1$

$x_{2} = \frac{- b - \sqrt{D} }{2a} = \frac{ - 11 - 27}{16} = \frac{- 38}{16} = \frac{- 19}{8} = -2\frac{3}{8}$

3) х² = 30 + х

х² - х - 30 = 0

D = b² - 4ac = 1 + 120 = 121 = 11² ⇒ 2 корня:

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

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

4) 16 - 8х + х² = 0

х² - 8х + 16 = 0

D = b² - 4ac = 64 - 64 = 0 ⇒ 1 корень:

$x = \frac{- b}{2a} = \frac{8}{2} = 4$

5) 5x² - 26x + 5 = 0

$D_{1} = (\frac{b}{2})^{2} - ac = 169 - 25 = 144 = 12^{2}$ ⇒ 2 корня:

$x_{1} = \frac{-\frac{b}{2} + \sqrt{D_{1} } }{a} = \frac{13 + 12}{5} = \frac{25}{5} = 5$

$x_{2} = \frac{-\frac{b}{2} - \sqrt{D_{1} } }{a} = \frac{13 - 12}{5} = \frac{1}{5} = 0,2$