1. Решите неравенство:
\(\left(\displaystyle \frac{1}{x^2-7x+12}+\frac{x-4}{3-x}\right)\cdot \sqrt{6x-x^2}\leq 0 \Leftrightarrow\)
\(\Leftrightarrow \left\{\begin{matrix}\left[
\begin{array}{ccc}\displaystyle \frac{1}{x^2-7x+12}+\frac{x-4}{3-x}\leq0,
\\6x-x^2=0,
\end{array}
\right.
\\ 6x-x^2\geq0; \end{matrix}\right. \Leftrightarrow \left[
\begin{array}{ccc}
\left\{\begin{matrix}
\displaystyle \frac{1}{(x-3)(x-4)}+\frac{(x-4)}{3-x}\leq0, \\ 6x-x^2\geq0, \end{matrix}\right.\\
6x-x^2=0 ,
\end{array}
\right. \Leftrightarrow x^2-7x+12=(x-3)(x-4).\)
\(\left[
\begin{array}{ccc}
\left\{\begin{matrix}
\displaystyle \frac{1}{(x-3)(x-4)}-\frac{(x-4)}{x-3}\leq0, \\ x(x-6)\leq0, \end{matrix}\right.\\
x=0, \\ x=6 ;
\end{array}
\right. \Leftrightarrow \left[
\begin{array}{ccc}
\left\{\begin{matrix}
\displaystyle \frac{1-(x-4)^2}{(x-3)(x-4)}\leq0, \\ 0 \leq x \leq 6, \end{matrix}\right.\\
x=0, \\ x=6; \end{array}
\right. \Leftrightarrow \left[
\begin{array}{ccc}
\left\{\begin{matrix}
\displaystyle \frac{(1-x+4)(1+x-4)}{(x-3)(x-4)}\leq0, \\ 0 \leq x \leq 6, \end{matrix}\right.\\
x=0, \\ x=6 ;
\end{array}
\right. \Leftrightarrow\)
\(\Leftrightarrow \left[
\begin{array}{ccc}
\left\{\begin{matrix}
\displaystyle \frac{(x-5)(x-3)}{(x-3)(x-4)}\geq0, \\ 0 \leq x \leq 6, \end{matrix}\right.\\
x=0, \\ x=6 ;
\end{array}
\right. \Leftrightarrow \left[
\begin{array}{ccc}
x < 3 ,\\
3 < x< 4 ,\\ 5\leq x\leq6. \end{array} \right. \)
Ответ: \(x\in [0;3) \cup (3;4) \cup [5;6].\)
2. Решите неравенство:
\( \sqrt{5-x} < \displaystyle \frac{\sqrt{x^3-7x^2+14x-5}}{\sqrt{x-1}} \Leftrightarrow\left\{\begin{matrix} 5-x \geq0 ,
\\x-1>0 ,
\\ x^3-7x^2+14x-5\geq0,
\\ 5-x < \displaystyle\frac{x^3-7x^2+14x-5}{x-1} ;\end{matrix}\right. \Leftrightarrow\)
\(\Leftrightarrow\left\{\begin{matrix}x\leq5 ,
\\x > 1 ,
\\x^3-7x^2+14x-5\geq0 ,
\\(5-x)(x-1)< x^3-7x^2+14x-5; \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
1< x \leq 5, \\ 5x-x^2-5+x < x^3-7x^2+14x-5 .\end{matrix}\right.\)
Поскольку \(x>1\), мы умножили на \(x-1\) обе части неравенства.
Так как \(5-x\geq 0, \, x-1 > 0\), то \((5-x)(x-1)\geq 0.\)
Получили, что \(0 \leq (5-x)(x-1)>x^3-7x^2+14x-5\), значит, \(x^3-7x^2+14x-5>0.\)
\(\left\{\begin{matrix}1 < x\leq 5 ,
\\x^3-6x^2+8x>0 ;\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} 1 < x\leq 5 ,
\\ x(x^2-6x+8)>0; \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} 1 < x\leq 5 ,
\\ x(x-2)(x-4) >0 .\end{matrix}\right. \)
Решив второе неравенство методом интервалов, получим:
\(\left\{\begin{matrix} \left[
\begin{array}{ccc}
0 < x < 2,\\
x>4,
\end{array}
\right.
\\ 1< x\leq 5; \end{matrix}\right. \Leftrightarrow \left[
\begin{array}{ccc}
1 < x < 2,\\
4< x\leq5. \end{array} \right.\)
Ответ: \(x\in (1;2)\cup(4;5].\)
3. Решите неравенство:
\(\left (x+\displaystyle \frac{3}{x}\right)\left (\displaystyle \frac{\sqrt{x^2-6x+9}-1}{\sqrt{5-x}-1}\right )^2 \geq 4 \left (\displaystyle \frac{\sqrt{x^2-6x+9}-1}{\sqrt{5-x}-1}\right )^2 \Leftrightarrow \left (x+\displaystyle \frac{3}{x}-4\right )\left (\displaystyle \frac{\sqrt{x^2-6x+9}-1}{\sqrt{5-x}-1}\right )^2 \geq0 \Leftrightarrow\)
\(\Leftrightarrow \displaystyle \frac{x^2-4x+3}{x}\cdot \left (\displaystyle \frac{\sqrt{(x-3)^2}-1}{\sqrt{5-x}-1}\right )^2\geq0 \Leftrightarrow \displaystyle \frac{(x-1)(x-3)}{x}\cdot \left (\displaystyle \frac{|x-3|-1}{\sqrt{5-x}-1}\right )^2\geq 0 \Leftrightarrow \)
\(\Leftrightarrow \left[
\begin{array}{ccc}
\displaystyle \frac{(x-1)(x-3)}{x}\geq0,\\ \displaystyle \frac{|x-3|-1}{\sqrt{5-x}-1}=0 ;\end{array}
\right. \Leftrightarrow \left\{\begin{matrix} \left[
\begin{array}{ccc} \displaystyle \frac{(x-1)(x-3)}{x}\geq0,
\\|x-3|-1=0, \end{array} \right.
\\ 5-x\geq0 ,\\ \sqrt{5-x}\ne1; \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} \left[
\begin{array}{ccc} \displaystyle \frac{(x-1)(x-3)}{x}\geq0,
\\x=4,\\x=2, \end{array} \right.
\\ x\leq5, \\ x\ne4 ;\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} \left[
\begin{array}{ccc} \displaystyle \displaystyle \frac{(x-1)(x-3)}{x}\geq0,
\\x=2, \end{array} \right.
\\ x\ne4,\\ x\leq5; \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} \left[
\begin{array}{ccc} 0 < x\leq1,
\\x\geq3, \\ x=2, \end{array}
\right.
\\ x\ne4,\\ x\leq5. \end{matrix}\right. \)
Ответ: \(x\in (0;1] \cup \left \{ 2 \right \} \cup [3;4) \cup (4;5].\)
