Question

Решить уравнение:
$log_{12 x^{2} -5x-2}(4x+1) leq 0$

Answer 1

Метод рационализации:

$log_{12x^2-5x-2} (4x+1)leq 0\\ODZ:; left { {{12x^2-5x-2ne 1 ,; 12x^2-5x-2 textgreater 0} atop {4x+1 textgreater 0}} right. ; left { {{xne frac{3}{4},xne -frac{1}{3},12x^2-5x-2 textgreater 0} atop {x textgreater -frac{1}{4}}} right.$

$12x^2-5x-2 textgreater 0; ,; D=25+96=121,\\x_1= frac{5-11}{24} =-frac{1}{4}; ,; x_2= frac{5+11}{24} =frac{2}{3}\\12(x+frac{1}{4})(x-frac{2}{3}) textgreater 0; ,; ; +++(-frac{1}{4})---(frac{2}{3})+++\\xin (-infty ,-frac{1}{4})cup (frac{2}{3},+infty ); ; Rightarrow\\ODZ:; xin ( frac{2}{3}, frac{3}{4})cup ( frac{3}{4},+infty )$

$star quad log_{h}f leq 0; ; Leftrightarrow ; ; (h-1)(f-1) leq 0\\(12x^2-5x-2-1)(4x+1-1) leq 0\\(12x^2-5x-3)cdot 4x leq 0\\12cdot (x-frac{3}{4})(x+frac{1}{3})cdot 4x leq 0\\Znaki:; ; ---[-frac{1}{3}]+++[, 0, ]---[frac{3}{4}]+++$

$xin (-infty ,-frac{1}{3}]cup [, 0,frac{3}{4}, ]$

$Otvet:xin (frac{2}{3},frac{3}{4})$