1. Giải bpt
a, 2 <= gttđ(x+1)
b, gttđ(x-2) < 3x-2
c, (2x+1).(x^2 + x - 30) > 0
2. Số nghiệm nguyên của bpt (x-2)/(x+1) > x/(x+3)
1. Giải bpt
a, 2 <= gttđ(x+1)
b, gttđ(x-2) < 3x-2
c, (2x+1).(x^2 + x - 30) > 0
2. Số nghiệm nguyên của bpt (x-2)/(x+1) > x/(x+3)
Đáp án:
$\begin{array}{l}
a)2 \le \left| {x + 1} \right|\\
\Rightarrow \left[ \begin{array}{l}
x + 1 \ge 2\\
x + 1 \le – 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le – 3
\end{array} \right.\\
b)\left| {x – 2} \right| < 3x – 2\\
\Rightarrow \left\{ \begin{array}{l}
3x – 2 > 0\\
– 3x + 2 < x – 2 < 3x – 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > \dfrac{2}{3}\\
1 < x
\end{array} \right.\\
\Rightarrow x > 1\\
c)\left( {2x + 1} \right)\left( {{x^2} + x – 30} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x + 1 > 0\\
{x^2} + x – 30 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x + 1 < 0\\
{x^2} + x – 30 < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > – \dfrac{1}{2}\\
\left( {x – 5} \right)\left( {x + 6} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < – \dfrac{1}{2}\\
\left( {x – 5} \right)\left( {x + 6} \right) < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > – \dfrac{1}{2}\\
x > 5/x < – 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x < – \dfrac{1}{2}\\
– 6 < x < 5
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x > 5\\
– 6 < x < – \dfrac{1}{2}
\end{array} \right.\\
2)\dfrac{{x – 2}}{{x + 1}} > \dfrac{x}{{x + 3}}\\
\Rightarrow \dfrac{{x + 1 – 3}}{{x + 1}} > \dfrac{{x + 3 – 3}}{{x + 3}}\\
\Rightarrow 1 – \dfrac{3}{{x + 1}} > 1 – \dfrac{3}{{x + 3}}\\
\Rightarrow \dfrac{3}{{x + 1}} < \dfrac{3}{{x + 3}}\\
\Rightarrow \dfrac{1}{{x + 1}} < \dfrac{1}{{x + 3}}\\
\Rightarrow \dfrac{{x + 3 – x – 1}}{{\left( {x + 1} \right)\left( {x + 3} \right)}} < 0\\
\Rightarrow \dfrac{2}{{\left( {x + 1} \right)\left( {x + 3} \right)}} < 0\\
\Rightarrow \left( {x + 1} \right)\left( {x + 3} \right) < 0\\
\Rightarrow – 3 < x < – 1
\end{array}$