№1 Решите неравенство
a)
0)\ \ or\ \ (x+3>0\ \ and\ \ x+6<0)\\\\(x<-3\ \ and\ \ x>-6)\ \ or\ \ (x>-3\ \ and\ \ x<-6)\\\\(-6<x<-3)\ \ or\ \ (no\ solutions)\\\\x\in(-6;\ -3)" alt="\frac{1}{3} x^2+3x+6<0\\\\x^2+9x+18<0\\\\x^2+2*x*\frac{9}{2}+(\frac{9}{2})^2-(\frac{9}{2})^2+18<0\\\\(x+\frac{9}{2})^2-\frac{81}{4}+18<0\\\\(x+4.5)^2-20.25+18<0\\\\(x+4.5)^2-2.25<0\\\\(x+4.5)^2-1.5^2<0\\\\(x+4.5-1.5)*(x+4.5+1.5)<0\\\\(x+3)*(x+6)<0\\\\(x+3<0\ \ and\ \ x+6>0)\ \ or\ \ (x+3>0\ \ and\ \ x+6<0)\\\\(x<-3\ \ and\ \ x>-6)\ \ or\ \ (x>-3\ \ and\ \ x<-6)\\\\(-6<x<-3)\ \ or\ \ (no\ solutions)\\\\x\in(-6;\ -3)" align="absmiddle" class="latex-formula">
Ответ: 
б)
0\ |*(-1)\\x^2-5x+16<0\\\\x^2-2*x*\frac{5}{2}+16<0\\\\x^2-2*x*\frac{5}{2}+(\frac{5}{2})^2-(\frac{5}{2})^2+16<0\\\\(x-\frac{5}{2})^2-\frac{25}{4}+16<0\\\\(x-\frac{5}{2})^2-6.25+16<0\\\\(x-\frac{5}{2})^2+9.75<0\\" alt="-x^2+5x-16>0\ |*(-1)\\x^2-5x+16<0\\\\x^2-2*x*\frac{5}{2}+16<0\\\\x^2-2*x*\frac{5}{2}+(\frac{5}{2})^2-(\frac{5}{2})^2+16<0\\\\(x-\frac{5}{2})^2-\frac{25}{4}+16<0\\\\(x-\frac{5}{2})^2-6.25+16<0\\\\(x-\frac{5}{2})^2+9.75<0\\" align="absmiddle" class="latex-formula">
Как можно видеть, неравенство не выполниться ни при каком значении 
, так как 
Ответ : решений нету
№2 Решите неравенство
применим метод интервалов
x\\\\x\in[-8;\ \frac{1}{3}]\cup[4;\ +\infty)" alt="x_1=4;\ \ x_2=\frac{1}{3};\ \ x_3=-8\\\\-----[-8]+++++[\frac{1}{3}]-----[4]++++>x\\\\x\in[-8;\ \frac{1}{3}]\cup[4;\ +\infty)" align="absmiddle" class="latex-formula">
Ответ: ![[-8;\ \frac{1}{3}]\cup[4;\ +\infty)](https://tex.z-dn.net/?f=%5B-8%3B%5C%20%5Cfrac%7B1%7D%7B3%7D%5D%5Ccup%5B4%3B%5C%20%2B%5Cinfty%29 "[-8;\ \frac{1}{3}]\cup[4;\ +\infty)")
№2 Решите систему неравенств
0\\ 2x+3 \geq 0 \end{cases}\end{equation*}\\\\\begin{equation*} \begin{cases} 5x^2-5x-4x+4>0\\ 2x \geq -3 \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} 5x(x-1)-4(x-1)>0\\ 2x \geq -3 \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} (5x-4)*(x-1)>0\\ x \geq -1.5 \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} 5*(x-\frac{4}{5})*(x-1)>0\\ x \geq -1.5 \end{cases}\end{equation*}\\" alt="\begin{equation*} \begin{cases} 5x^2-9x+4>0\\ 2x+3 \geq 0 \end{cases}\end{equation*}\\\\\begin{equation*} \begin{cases} 5x^2-5x-4x+4>0\\ 2x \geq -3 \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} 5x(x-1)-4(x-1)>0\\ 2x \geq -3 \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} (5x-4)*(x-1)>0\\ x \geq -1.5 \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} 5*(x-\frac{4}{5})*(x-1)>0\\ x \geq -1.5 \end{cases}\end{equation*}\\" align="absmiddle" class="latex-formula">
0\\ x +1.5\geq0 \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} +++++++++(0.8)----(1)++++>x\\ ----[-1.5]++++++++++++++>x \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} x\in(-\infty;\ 0.8)\cup(1;\ +\infty)\\ x\in[-1.5;\ +\infty) \end{cases}\\\end{equation*}\\ x\in[-1.5;\ 0.8)\cup(1;\ +\infty)\\" alt="\begin{equation*} \begin{cases} (x-0.8)*(x-1)>0\\ x +1.5\geq0 \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} +++++++++(0.8)----(1)++++>x\\ ----[-1.5]++++++++++++++>x \end{cases}\end{equation*}\\\begin{equation*} \begin{cases} x\in(-\infty;\ 0.8)\cup(1;\ +\infty)\\ x\in[-1.5;\ +\infty) \end{cases}\\\end{equation*}\\ x\in[-1.5;\ 0.8)\cup(1;\ +\infty)\\" align="absmiddle" class="latex-formula">
Ответ: 