Question

помогите решить неравенство (х+1)log x (3x+7)≥0

Answer 1

$(x+1)log_x(3x+7) \geq 0 <=> \left \{ {{(x+1)(x-1)(3x+7-1) \geq 0} \atop {x > 0}}\atop {x \neq 1; 3x+7 > 0}} \right\\(x+1)(x-1)(3x+7-1) \geq 0\\(x+1)(x-1)(3x+6) \geq 0\\(x+1)(x-1)(x+2) \geq 0\\-------[-2]++++[-1]------[1]+++++>x\\x \in [-2;-1]  [1;+\infty)\\x>0\\x\neq 1\\x > -\frac{7}{3} \\x\in(1;+\infty)\\Answer: x\in(1;+\infty)$