Метод рационализации: 
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0} \atop {6x^2+5x>0,\; 6x^2+5x\ne 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 \{ {{x\in (-\infty ;\, 0,5)\cup (1,+\infty )\qquad \qquad } \atop {x\in (-\infty ;-\frac{5}{6})\cup (0,+\infty )\; ,\; x\ne -1\; ,\; x\ne \frac{1}{6}}} \right. \\\\x\in (-\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" alt="log_{6x^2+5x}(2x^2-3x+1)\geq 0\\\\ODZ:\; \; \left \{ {{2x^2-3x+1>0} \atop {6x^2+5x>0,\; 6x^2+5x\ne 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 \{ {{x\in (-\infty ;\, 0,5)\cup (1,+\infty )\qquad \qquad } \atop {x\in (-\infty ;-\frac{5}{6})\cup (0,+\infty )\; ,\; x\ne -1\; ,\; x\ne \frac{1}{6}}} \right. \\\\x\in (-\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" align="absmiddle" class="latex-formula">
![x(2x-3)\cdot 6\cdot (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\, ]+++\\\\x\in (-\infty ,-1)\cup (0,\frac{1}{6})\cup [\, 1,5\, ;\, +\infty )\\\\\left \{ {{x\in (-\infty ,-1)\cup (0,\frac{1}{6})\cup [\, 1,5\, ;\, +\infty )} \atop {x\in (-\infty ,-1)\cup (-1,-\frac{5}{6})\cup (\frac{1}{6},\frac{1}{2})\cup (1,+\infty )}} \right. \; \; \Rightarrow \; \; \underline {x\in (-\infty ,-1)\cup [\, 1,5\, ;\, +\infty )}](https://tex.z-dn.net/?f=x%282x-3%29%5Ccdot%206%5Ccdot%20%28x%2B1%29%28x-%5Cfrac%7B1%7D%7B6%7D%29%5Cgeq%200%5C%5C%5C%5Cx_1%3D0%5C%3B%20%2C%5C%3B%20x_2%3D1%2C5%5C%3B%20%2C%5C%3B%20x_3%3D-1%5C%3B%20%2C%5C%3B%20x_4%3D%5Cfrac%7B1%7D%7B6%7D%5C%5C%5C%5C%2B%2B%2B%28-1%29---%280%29%2B%2B%2B%28%5Cfrac%7B1%7D%7B6%7D%29---%5B%5C%2C%201%2C5%5C%2C%20%5D%2B%2B%2B%5C%5C%5C%5Cx%5Cin%20%28-%5Cinfty%20%2C-1%29%5Ccup%20%280%2C%5Cfrac%7B1%7D%7B6%7D%29%5Ccup%20%5B%5C%2C%201%2C5%5C%2C%20%3B%5C%2C%20%2B%5Cinfty%20%29%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7Bx%5Cin%20%28-%5Cinfty%20%2C-1%29%5Ccup%20%280%2C%5Cfrac%7B1%7D%7B6%7D%29%5Ccup%20%5B%5C%2C%201%2C5%5C%2C%20%3B%5C%2C%20%2B%5Cinfty%20%29%7D%20%5Catop%20%7Bx%5Cin%20%28-%5Cinfty%20%2C-1%29%5Ccup%20%28-1%2C-%5Cfrac%7B5%7D%7B6%7D%29%5Ccup%20%28%5Cfrac%7B1%7D%7B6%7D%2C%5Cfrac%7B1%7D%7B2%7D%29%5Ccup%20%281%2C%2B%5Cinfty%20%29%7D%7D%20%5Cright.%20%5C%3B%20%5C%3B%20%5CRightarrow%20%5C%3B%20%5C%3B%20%5Cunderline%20%7Bx%5Cin%20%28-%5Cinfty%20%2C-1%29%5Ccup%20%5B%5C%2C%201%2C5%5C%2C%20%3B%5C%2C%20%2B%5Cinfty%20%29%7D "x(2x-3)\cdot 6\cdot (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\, ]+++\\\\x\in (-\infty ,-1)\cup (0,\frac{1}{6})\cup [\, 1,5\, ;\, +\infty )\\\\\left \{ {{x\in (-\infty ,-1)\cup (0,\frac{1}{6})\cup [\, 1,5\, ;\, +\infty )} \atop {x\in (-\infty ,-1)\cup (-1,-\frac{5}{6})\cup (\frac{1}{6},\frac{1}{2})\cup (1,+\infty )}} \right. \; \; \Rightarrow \; \; \underline {x\in (-\infty ,-1)\cup [\, 1,5\, ;\, +\infty )}")