Question

помогите решить задания

Answer 1

Ответ:

7. {3;9} (ответ 1)

8. 1/8 (ответ 2)

9. 0,5 (ответ 3)

Пошаговое объяснение:

7.

$A = \{1; 3; 5; 7; 9\}; \quad B = \{3;6;9;12\} \\ \small {A \setminus B = \{ \cancel{1}; 3; \cancel{5};\cancel{ 7};9 \}\setminus \{3;\cancel{6};{9};\cancel{12}\} = \{3;9 \}}$

8.

$\lim_{x \to \infty} \left( \frac{ {x}^{3} - x + 2 }{8{x}^{3} + x + 1 } \right) = \\ = \lim_{x \to \infty} \left( \frac{1}{8} \cdot \frac{ 8{x}^{3} - 8x + 16 }{8{x}^{3} + x + 1 } \right) = \\ = \frac{1}{8} \lim_{x \to \infty} \left( \frac{8{x}^{3} + x + 1 - 9x + 15 }{8{x}^{3} + x + 1 } \right) = \\ = \tfrac{1}{8} \lim_{x \to \infty} \left( \tfrac{8{x}^{3} + x + 1 }{8{x}^{3} + x + 1 } - \tfrac{9x }{8{x}^{3} + x + 1 } + \tfrac{ 15 }{8{x}^{3} + x + 1 } \right) = \\ \small{ = \frac{1}{8}\left( \lim_{x \to \infty} \left( \tfrac{8{x}^{3} + x + 1 }{8{x}^{3} + x + 1 }\right) - \lim_{x \to \infty} \left( \tfrac{9x }{8{x}^{3} + x + 1 }\right) + \lim_{x \to \infty} \left(\tfrac{ 15 }{8{x}^{3} + x + 1 } \right)\right) = } \\ = \frac{1}{8} \times (1 - 0 + 0) = \frac{1}{8}$

9.

$\lim_{x \to \ 3} \left(\dfrac{ {x}^{2} - 3x } { {x}^{2} - 9 }\right) = \left[{\dfrac{0}{0}} \right] = \\ = \lim_{x \to \ 3} \left(\frac{ x \cancel{(x - 3)}} {\cancel{(x - 3)}(x + 3)}\right) = \\ =\lim_{x \to \ 3} \left(\frac{ x} {x + 3}\right) ={\dfrac{3}{3 + 3}} = \frac{1}{2} = 0.5$