Question

Розв'яжіть нерівність.
2; 3; 6; 7; 8.
СРОЧНО. ДАЮ 100Б.​

Answer 1

2)

$\frac{x - 9}{x + 11} > 0 \\ (x - 9)(x + 11) > 0 \\ + + + ( - 11) - - - (9) + + + \\ x \: \epsilon \: ( - \propto; \: - 11)U(9; \: + \propto)$

3)

$\frac{x - 3.2}{x - 4.8} \geqslant 0 \\ \\ \left \{ {{( x- 3.2)(x - 4.8) \geqslant 0} \atop {x \neq4.8}} \right. \\ \\ + + + [3.2] - - - (4.8) + + + \\ x \: \epsilon \:( - \propto; \: 3.2] U(4.8; \: + \propto)$

6)

$\frac{3x + 1.8}{1.5 - 5x} \leqslant 0 \\ \frac{x + 0.6}{x - 0.3} \geqslant 0 \\ \\\left \{ {{(x + 0.6)(x - 0.3) \geqslant 0} \atop {x\neq0.3 }} \right. \\ \\ + + + [ -0.6] - - - (0.3) + + + \\ x \: \epsilon \: ( - \propto; \: - 0.6]U(0.3; \: + \propto)$

7)

$\frac{(x + 13)(x + 2)}{x - 13} \geqslant 0 \\ \\ \left \{ {{(x + 13)(x + 2)(x - 13) \geqslant 0} \atop {x\neq13 }} \right. \\ \\ - - - [ - 13] + + + [ - 2] - - - (13) + + + \\ x \: \epsilon \: [ - 13; \: - 2]U(13; \: + \propto)$

8)

$\frac{x - 3.5}{(x + 6)(x - 12)} \leqslant 0 \\ \\ \left \{ {{(x - 3.5)(x + 6)(x - 12) \leqslant 0} \atop {x\neq - 6 \: \: \: \: \: \: \: x\neq 12}} \right. \\ \\ - - - ( - 6) + + + [3.5] - - - (12) + + + \\ x \: \epsilon \: ( - \propto; \: - 6)U[3.5; \: 12)$