R - множество действительных чисел.
![\frac{3x-7}{6} \geq \frac{5-6x}{4} |*12\\2(3x-7)\geq 3(5-6x)\\6x+18x\geq 14+15\\24x\geq 29|:24\\x\geq \frac{29}{24} \\x\in [\frac{29}{24};+\infty) \frac{3x-7}{6} \geq \frac{5-6x}{4} |*12\\2(3x-7)\geq 3(5-6x)\\6x+18x\geq 14+15\\24x\geq 29|:24\\x\geq \frac{29}{24} \\x\in [\frac{29}{24};+\infty)](https://tex.z-dn.net/?f=%5Cfrac%7B3x-7%7D%7B6%7D%20%5Cgeq%20%5Cfrac%7B5-6x%7D%7B4%7D%20%7C%2A12%5C%5C2%283x-7%29%5Cgeq%203%285-6x%29%5C%5C6x%2B18x%5Cgeq%2014%2B15%5C%5C24x%5Cgeq%2029%7C%3A24%5C%5Cx%5Cgeq%20%5Cfrac%7B29%7D%7B24%7D%20%5C%5Cx%5Cin%20%5B%5Cfrac%7B29%7D%7B24%7D%3B%2B%5Cinfty%29)
Ответ: ![[\frac{29}{24};+\infty) [\frac{29}{24};+\infty)](https://tex.z-dn.net/?f=%5B%5Cfrac%7B29%7D%7B24%7D%3B%2B%5Cinfty%29)
![\frac{x-1}{4} +\frac{x+3}{2}<1-\frac{x}{6}|\cdot 12\\3(x-1)+6(x+3)<12-2x\\3x+6x+2x<3-18+12\\11x<-3|:11\\x<-\frac{3}{11}\\x\in (-\infty;-\frac{3}{11}) \frac{x-1}{4} +\frac{x+3}{2}<1-\frac{x}{6}|\cdot 12\\3(x-1)+6(x+3)<12-2x\\3x+6x+2x<3-18+12\\11x<-3|:11\\x<-\frac{3}{11}\\x\in (-\infty;-\frac{3}{11})](https://tex.z-dn.net/?f=%5Cfrac%7Bx-1%7D%7B4%7D%20%2B%5Cfrac%7Bx%2B3%7D%7B2%7D%3C1-%5Cfrac%7Bx%7D%7B6%7D%7C%5Ccdot%2012%5C%5C3%28x-1%29%2B6%28x%2B3%29%3C12-2x%5C%5C3x%2B6x%2B2x%3C3-18%2B12%5C%5C11x%3C-3%7C%3A11%5C%5Cx%3C-%5Cfrac%7B3%7D%7B11%7D%5C%5Cx%5Cin%20%28-%5Cinfty%3B-%5Cfrac%7B3%7D%7B11%7D%29)
Ответ: ![(-\infty;-\frac{3}{11}) (-\infty;-\frac{3}{11})](https://tex.z-dn.net/?f=%28-%5Cinfty%3B-%5Cfrac%7B3%7D%7B11%7D%29)
\frac{16}{6}\\x\in (2\frac{2}{3};+\infty)" alt="\tt (x-3)(x-6)<(x-1)(x-2)\\x^2-3x-6x+18<x^2-x-2x+2\\-6x<-16|:(-6)\\x>\frac{16}{6}\\x\in (2\frac{2}{3};+\infty)" align="absmiddle" class="latex-formula">
Ответ: ![(2\frac{2}{3};+\infty) (2\frac{2}{3};+\infty)](https://tex.z-dn.net/?f=%282%5Cfrac%7B2%7D%7B3%7D%3B%2B%5Cinfty%29)