Question

(х+2)(х-3)/х(3х+2)≤0 решите неравенство помогите пожалуйста дам много баллов дам 80​

Answer 1

а)

${x}^{2} - 6x + 8 > 0 \\ \\ {x {}^{2} - 6x + 8}^{} = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ x_{1} + x_{2} = 6 \\ x_{1} x_{2} = 8\\ x_{1} = 4\\ x_{2} = 2 \\ {x}^{2} - 6x + 8 = (x - 2)(x - 4) \\ \\ (x - 2)(x - 4) > 0 \\ + + + (2) - - - (4) + + + \\ otvet \: \: \: x \: \epsilon \: ( - \propto; \: 2)U(4; \: + \propto)$

б)

$\frac{(x + 2)(x - 3)}{x(3x + 2)} \leqslant 0 \\ \frac{(x + 2)(x - 3)}{x(x + \frac{2}{3} )} \leqslant 0 \\ \\ \left \{ {{x(x + 2)(x - 3)(x + \frac{2}{3} ) \leqslant 0} \atop {x\neq0 \: \: \: and \: \: \: x\neq - \frac{2}{3} }} \right. \\ + + [ - 2] - - ( - \frac{2}{3} ) + + (0) - - [3] + + \\ otvet \: \: \: x \: \epsilon\: [ - 2; \: - \frac{2}{3})U(0; \: 3]$