Bài 1: -4x^2+x-3 >0
b) 3x^2+x+5>0
c) -9-2x-7x^2<0
Bài 2: |3x^2-4x-7|<2
Bài 3: Cho f(x)=(m-2)x+2(m-1)-2.Tìm m để f(x)>0
Với mọi x thuộc R
Bài 1: -4x^2+x-3 >0
b) 3x^2+x+5>0
c) -9-2x-7x^2<0
Bài 2: |3x^2-4x-7|<2
Bài 3: Cho f(x)=(m-2)x+2(m-1)-2.Tìm m để f(x)>0
Với mọi x thuộc R
Đáp án:
$\begin{array}{l}
1)a) – 4{x^2} + x – 3 > 0\\
\Rightarrow 4{x^2} – x + 3 > 0\\
\Rightarrow 4{x^2} – 2.2x.\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{47}}{{16}} > 0\\
\Rightarrow {\left( {2x – \dfrac{1}{4}} \right)^2} + \dfrac{{47}}{{16}} > 0\left( {tm} \right)\\
Vậy\,x \in R\\
b)3{x^2} + x + 5 > 0\\
\Rightarrow {x^2} + \dfrac{1}{3}.x + \dfrac{5}{3} > 0\\
\Rightarrow {x^2} + 2.x.\dfrac{1}{6} + \dfrac{1}{{36}} + \dfrac{{59}}{{36}} > 0\\
\Rightarrow {\left( {x + \dfrac{1}{6}} \right)^2} + \dfrac{{59}}{{36}} > 0\left( {tm} \right)\\
Vậy\,x \in R\\
c) – 9 – 2x – 7{x^2} < 0\\
\Rightarrow 7{x^2} + 2x + 9 > 0\\
\Rightarrow {x^2} + 2.\dfrac{1}{7}.x + \dfrac{9}{7} > 0\\
\Rightarrow {\left( {x + \dfrac{1}{7}} \right)^2} + \dfrac{{62}}{{49}} > 0\left( {tm} \right)\\
Vậy\,x \in R\\
2)\left| {3{x^2} – 4x – 7} \right| < 2\\
\Rightarrow – 2 < 3{x^2} – 4x – 7 < 2\\
\Rightarrow \left\{ \begin{array}{l}
3{x^2} – 4x – 5 > 0\\
3{x^2} – 4x – 9 < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > \dfrac{{2 + \sqrt {19} }}{3}\\
x < \dfrac{{2 – \sqrt {19} }}{3}
\end{array} \right.\\
\left[ {\dfrac{{2 – \sqrt {31} }}{3} < x < \dfrac{{2 + \sqrt {31} }}{3}} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{{2 – \sqrt {31} }}{3} < x < \dfrac{{2 – \sqrt {19} }}{3}\\
\dfrac{{2 + \sqrt {19} }}{3} < x < \dfrac{{2 + \sqrt {31} }}{3}
\end{array} \right.
\end{array}$