[tex]f(x) = \sqrt{x - 9} \\ f'(x) = \frac{1}{2 \sqrt{x - 9} } [/tex]
[tex]3 {x}^{2} - 2x + 7 = 0 \\ a = 3 \\ b = - 2 \\ c =7 \\ D = {b}^{2} - 4ac = ( - 2) {}^{2} - 4 \times 3 \times 7 = 4 - 84 = - 80 \\ D < 0[/tex]
Ответ: нет корней
[tex] {x}^{3} - 2x < (x + 2)( {x}^{2} - 2x + 4) \\ {x}^{3} - 2x < {x}^{3} + 8 \\ {x}^{3} - 2x - {x}^{3} < 8 \\ -2 x < 8 \: \: | \div ( - 2) \\ x > - 4 \\ x \: \epsilon \: ( - 4; \: + \propto)[/tex]
Ответ: А)
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
1.
[tex]f(x) = \sqrt{x - 9} \\ f'(x) = \frac{1}{2 \sqrt{x - 9} } [/tex]
2.
[tex]3 {x}^{2} - 2x + 7 = 0 \\ a = 3 \\ b = - 2 \\ c =7 \\ D = {b}^{2} - 4ac = ( - 2) {}^{2} - 4 \times 3 \times 7 = 4 - 84 = - 80 \\ D < 0[/tex]
Ответ: нет корней
3.
[tex] {x}^{3} - 2x < (x + 2)( {x}^{2} - 2x + 4) \\ {x}^{3} - 2x < {x}^{3} + 8 \\ {x}^{3} - 2x - {x}^{3} < 8 \\ -2 x < 8 \: \: | \div ( - 2) \\ x > - 4 \\ x \: \epsilon \: ( - 4; \: + \propto)[/tex]
Ответ: А)