1) Giải BPT
$-1< \frac{10x^2-3x-2}{(x-1)(2-x)}< 1$
2) Tìm m sao cho
$1\leq \frac{3x^2-mx+5}{3x^2-x+1}\leq 6$ với mọi x
3) Giải BPT
$\left [ x \right ]\left \{ x \right \}< x-1$
$\left [ x \right ]\left \{ x \right \}< x-1$
Started By yeutoan11, 02-03-2012 - 17:26
#1
Posted 02-03-2012 - 17:26
Dựng nước lấy việc học làm đầu. Muốn thịnh trị lấy nhân tài làm gốc.
NGUYỄN HUỆ
Nguyễn Trần Huy
Tự hào là thành viên VMF
NGUYỄN HUỆ
Nguyễn Trần Huy
Tự hào là thành viên VMF
#2
Posted 02-03-2012 - 17:55
Bài 3:
\[\begin{array}{l}
\left[ x \right]\left\{ x \right\} < x - 1 \Leftrightarrow \left[ x \right] + \left\{ x \right\} - \left[ x \right]\left\{ x \right\} - 1 > 0 \\
\left. \begin{array}{l}
\Leftrightarrow \left( {\left[ x \right] - 1} \right)\left( {1 - \left\{ x \right\}} \right) > 0 \\
1 - \left\{ x \right\} > 0 \\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}
1 - \left\{ x \right\} > 0 \\
\left[ x \right] - 1 > 0 \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
0 \le \left\{ x \right\} < 1 \\
1 < \left[ x \right] \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
0 \le \left\{ x \right\} < 1 \\
x > 1 \\
\end{array} \right. \Leftrightarrow x \ge 2 \\
\end{array}\]
\[\begin{array}{l}
\left[ x \right]\left\{ x \right\} < x - 1 \Leftrightarrow \left[ x \right] + \left\{ x \right\} - \left[ x \right]\left\{ x \right\} - 1 > 0 \\
\left. \begin{array}{l}
\Leftrightarrow \left( {\left[ x \right] - 1} \right)\left( {1 - \left\{ x \right\}} \right) > 0 \\
1 - \left\{ x \right\} > 0 \\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}
1 - \left\{ x \right\} > 0 \\
\left[ x \right] - 1 > 0 \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
0 \le \left\{ x \right\} < 1 \\
1 < \left[ x \right] \\
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
0 \le \left\{ x \right\} < 1 \\
x > 1 \\
\end{array} \right. \Leftrightarrow x \ge 2 \\
\end{array}\]
- hoangtrong2305 likes this
Luôn yêu để sống, luôn sống để học toán, luôn học toán để yêu!!!
$$\text{LOVE}\left( x \right)|_{x = \alpha}^\Omega = + \infty $$
I'm still there everywhere.
$$\text{LOVE}\left( x \right)|_{x = \alpha}^\Omega = + \infty $$
I'm still there everywhere.
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