Question

Помогите решить логарифмическое неравенство

$log_{6x^{2} +5x} (2x^{2} -3x+1)geq 0$

Answer 1

Метод рационализации: $log_{g}fvee 0; ; Leftrightarrow ; ; (g-1)(f-1)vee 0$ .

$log_{6x^2+5x}(2x^2-3x+1)geq 0\\ODZ:; ; left { {{2x^2-3x+1>0} atop {6x^2+5x>0,; 6x^2+5xne 1}} right. ; left { {{2(x-1)(x-0,5)>0} atop {x(6x+5)>0,; 6(x+1)(x-frac{1}{6})ne 0}} right. \\left { {{xin (-infty ;, 0,5)cup (1,+infty )qquad qquad } atop {xin (-infty ;-frac{5}{6})cup (0,+infty ); ,; xne -1; ,; xne frac{1}{6}}} right. \\xin (-infty ,-1)cup (-1,-frac{5}{6})cup (frac{1}{6},frac{1}{2})cup (1,+infty )\\\(2x^2-3x+1-1)(6x^2+5x-1)geq 0$

$x(2x-3)cdot 6cdot (x+1)(x-frac{1}{6})geq 0\\x_1=0; ,; x_2=1,5; ,; x_3=-1; ,; x_4=frac{1}{6}\\+++(-1)---(0)+++(frac{1}{6})---[, 1,5, ]+++\\xin (-infty ,-1)cup (0,frac{1}{6})cup [, 1,5, ;, +infty )\\left { {{xin (-infty ,-1)cup (0,frac{1}{6})cup [, 1,5, ;, +infty )} atop {xin (-infty ,-1)cup (-1,-frac{5}{6})cup (frac{1}{6},frac{1}{2})cup (1,+infty )}} right. ; ; Rightarrow ; ; underline {xin (-infty ,-1)cup [, 1,5, ;, +infty )}$