previous arrow
next arrow
Slider

Решение домашнего задания от 13 мая

1. Решите неравенство

(\frac{1}{x^2-7x+12}+\frac{x-4}{3-x})\cdot \sqrt{6x-x^2}\leq 0 \Leftrightarrow

\Leftrightarrow \left\{\begin{matrix}\left[\begin{array}{ccc}\frac{1}{x^2-7x+12}+\frac{x-4}{3-x}\leq0 \hfill\\6x-x^2=0 \hfill\end{array}\right.\\ 6x-x^2\geq0 \hfill\end{matrix}\right. \Leftrightarrow \left[\begin{array}{ccc}\left\{\begin{matrix}\frac{1}{(x-3)(x-4)}+\frac{(x-4)}{3-x}\leq0 \hfill\\ 6x-x^2\geq0 \hfill\end{matrix}\right.\\6x-x^2=0 \hfill\end{array}\right. \Leftrightarrow

x^2-7x+12=(x-3)(x-4)

\left[\begin{array}{ccc}\left\{\begin{matrix}\frac{1}{(x-3)(x-4)}-\frac{(x-4)}{x-3}\leq0\\ x(x-6)\leq0 \hfill\end{matrix}\right.\\x=0 \hfill \\ x=6 \hfill\end{array}\right. \Leftrightarrow \left[\begin{array}{ccc}\left\{\begin{matrix}\frac{1-(x-4)^2}{(x-3)(x-4)}\leq0\\ 0 \leq x \leq 6 \hfill\end{matrix}\right.\\x=0 \hfill \\ x=6 \hfill\end{array}\right. \Leftrightarrow

\left[\begin{array}{ccc}\left\{\begin{matrix}\frac{(1-x+4)(1+x-4)}{(x-3)(x-4)}\leq0\\ 0 \leq x \leq 6 \hfill\end{matrix}\right.\\x=0 \hfill \\ x=6 \hfill\end{array}\right. \Leftrightarrow \left[\begin{array}{ccc}\left\{\begin{matrix}\frac{(x-5)(x-3)}{(x-3)(x-4)}\geq0\\ 0 \leq x \leq 6\end{matrix}\right.\\x=0 \hfill \\ x=6 \hfill\end{array}\right. \Leftrightarrow \left[\begin{array}{ccc}x \, \textless \, 3 \hfill\\3\, \textless \, x\, \textless \, \, 4 \hfill\\ 5\leq x\leq6 \hfill\end{array}\right.

Ответ: x\in [0;3) \cup (3;4) \cup [5;6].

2. Решите неравенство

\sqrt{5-x} \, \textless \, \frac{\sqrt{x^3-7x^2+14x-5}}{\sqrt{x-1}} \Leftrightarrow\left\{\begin{matrix} 5-x \geq0 \hfill\\x-1\, \textgreater \, 0 \hfill\\ x^3-7x^2+14x-5\geq0 \hfill\\ 5-x \, \textless \, \frac{x^3-7x^2+14x-5}{x-1}\end{matrix}\right. \Leftrightarrow

\left\{\begin{matrix}x\leq5 \hfill\\x \, \textgreater \, 1 \hfill\\x^3-7x^2+14x-5\geq0 \hfill\\(5-x)(x-1)\, \textless \, x^3-7x^2+14x-5\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}1\, \textless \, x \leq 5 \hfill\\ 5x-x^2-5+x \, \textless \, x^3-7x^2+14x-5 \hfill\end{matrix}\right.

Поскольку x\, \textgreater \, 1, мы умножили на x-1 обе части неравенства.

Так как 5-x\geq 0, \, \, x-1 \, \textgreater \, 0, то (5-x)(x-1)\geq 0.

Получили, что 0 \leq (5-x)(x-1)\, \textless \, x^3-7x^2+14x-5, значит, x^3-7x^2+14x-5\, \textgreater \, 0.

\left\{\begin{matrix}1 \, \textless \, x\leq 5 \hfill\\x^3-6x^2+8x\, \textgreater \,0\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} 1 \, \textless \, x\leq 5 \hfill\\ x(x^2-6x+8)\, \textgreater \, 0\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} 1 \, \textless \, x\leq 5 \hfill\\ x(x-2)(x-4) \, \textgreater \, 0\end{matrix}\right.

Решив второе неравенство методом интервалов, получим

\left\{\begin{matrix} \left[\begin{array}{ccc}0 \, \textless \, x \, \textless \,2\\x\,\textgreater \, 4\end{array}\right.\\ 1\, \textless \, x\leq 5\end{matrix}\right. \Leftrightarrow \left[\begin{array}{ccc}1 \, \textless \, x \, \textless \,2\\4\,\textless \, x\leq5\end{array}\right.

Ответ: x\in (1;2)\cup(4;5].

3. Решите неравенство

(x+\frac{3}{x})(\frac{\sqrt{x^2-6x+9}-1}{\sqrt{5-x}-1})^2 \geq 4 (\frac{\sqrt{x^2-6x+9}-1}{\sqrt{5-x}-1})^2 \Leftrightarrow

\Leftrightarrow (x+\frac{3}{x}-4)(\frac{\sqrt{x^2-6x+9}-1}{\sqrt{5-x}-1})^2 \geq0 \Leftrightarrow

\Leftrightarrow \frac{x^2-4x+3}{x}\cdot (\frac{\sqrt{(x-3)^2}-1}{\sqrt{5-x}-1})^2\geq0 \Leftrightarrow \frac{(x-1)(x-3)}{x}\cdot (\frac{|x-3|-1}{\sqrt{5-x}-1})^2\geq 0 \Leftrightarrow

\Leftrightarrow \left[\begin{array}{ccc}\frac{(x-1)(x-3)}{x}\geq0\\ \\ \frac{|x-3|-1}{\sqrt{5-x}-1}=0\end{array}\right. \Leftrightarrow \left\{\begin{matrix} \left[\begin{array}{ccc} \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

\Leftrightarrow \left\{\begin{matrix} \left[\begin{array}{ccc} \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} \frac{(x-1)(x-3)}{x}\geq0\\x=2\end{array}\right.\\ x\ne4\\ x\leq5\end{matrix}\right. \Leftrightarrow

\Leftrightarrow \left\{\begin{matrix} \left[\begin{array}{ccc} 0 \, \textless \, 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]