Question

Решите неравенство методом интервалов. (9x^2-4)(16-x^2)(2x^2+3)>0

Answer 1

$(9x^2-4)(16-x^2)(2x^2+3) textgreater 0\\(3x-2)(3x+2)(4-x)(4+x)(2x^2+3) textgreater 0\\(3x-2)(3x+2)(x-4)(x+4)(2x^2+3) textless 0\\2x^2+3 textgreater 0; ; pri; ; xin R\\+++(-4)---(-frac{2}{3})+++(frac{2}{3})---(4)+++\\xin (-4,-frac{2}{3})cup (frac{2}{3},4)$