Đề kiểm tra Dấu tam thức bậc hai lớp 10 (có lời giải) - Đề 2

Ta có: \(f(x) = 0 \Leftrightarrow  - {x^2} + 5x - 4 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = 4\end{array} \right.\).

Bảng xét dấu:

\(f\left( x \right) =  - {x^2} + 5x - 4 > 0 \Leftrightarrow 1 < x < 4\) hay \(x \in \left( {1;4} \right)\).

Ta có \(\Delta ' = 0,\,\,a < 0\) suy ra \( - 4{x^2} + 12x - 9 < 0,\,\,\forall x \in \mathbb{R}\backslash \left\{ {\frac{3}{2}} \right\}\).

Xét dấu \(f(x) = {x^2} + x - 6\)

Vậy \({x^2} + x - 6 \le 0 \Leftrightarrow x \in \left[ { - 3;2} \right]\)

Ta có: \[f(x) \ge 0,\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}a > 0\\\Delta  \le 0\end{array} \right.\].

Ta có: \(f\left( x \right) > 0\,,\,\,\forall x \in \mathbb{R}\) \( \Leftrightarrow \,\,\,\left\{ \begin{array}{l}a > 0\\\Delta  < 0\end{array} \right.\) \( \Leftrightarrow \,\,\,\left\{ \begin{array}{l}1 > 0\\{\left( {m + 2} \right)^2} - 4\left( {2m + 9} \right) < 0\end{array} \right.\) \( \Leftrightarrow \,\,\,{m^2} - 4m - 32 < 0\)

ĐKXĐ: \[\left\{ \begin{array}{l}x - 2 \ne 0\\x + 1 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 2\\x \ne  - 1\end{array} \right.\].

Ta có:

\[\frac{x}{{x - 2}} - \frac{2}{{x + 1}} \ge 1 \Leftrightarrow \frac{{{x^2} + x - 2x + 4}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} - \frac{{(x - 2)(x + 1)}}{{(x - 2)(x + 1)}} \ge 0\]

\[ \Leftrightarrow \frac{{{x^2} - x + 4 - {x^2} + x + 2}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} \ge 0 \Leftrightarrow \frac{6}{{\left( {x - 2} \right)\left( {x + 1} \right)}} \ge 0 \Leftrightarrow {x^2} - x - 2 > 0\]

\[ \Leftrightarrow \left[ \begin{array}{l}x <  - 1\\x > 2\end{array} \right.\].

Nhận xét: \({x^2} + 1 > 0\,,\,\,\forall x \in \mathbb{R} \Rightarrow f\left( x \right) < 0 \Leftrightarrow {x^2} - x - 6 < 0 \Leftrightarrow  - 2 < x < 3\).

Vậy \(x \in \left\{ { - 1;0;1;2} \right\}\). Có \(4\) giá trị.