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NV
11 tháng 2 2020

\(\left(2x+1\right)\left(x-1\right)>0\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -\frac{1}{2}\end{matrix}\right.\)

\(\left(3x+1\right)\left(x-5\right)\left(-4x+5\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-\frac{1}{3}\\\frac{5}{4}\le x\le5\end{matrix}\right.\)

\(\frac{x+2}{x-2}\le\frac{3x+1}{2x-1}\Leftrightarrow\frac{3x+1}{2x-1}-\frac{x+2}{x-2}\ge0\)

\(\Leftrightarrow\frac{x^2-8x}{\left(2x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\frac{x\left(x-8\right)}{\left(2x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\left[{}\begin{matrix}x\le0\\\frac{1}{2}< x< 2\\x\ge8\end{matrix}\right.\)

c: \(\Leftrightarrow\left\{{}\begin{matrix}4x+3>=0\\\left(x+2-4x-3\right)\left(x+2+4x+3\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{4}\\\left(-3x-1\right)\left(5x+5\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{4}\\\left(3x+1\right)\left(x+1\right)>0\end{matrix}\right.\)

\(\Leftrightarrow x>-\dfrac{1}{3}\)

d: \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-2< 0\\2x+1>=0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-2>=0\\\left(2x+1-3x+2\right)\left(2x+1+3x-2\right)>=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{2}{3}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(-x+3\right)\left(5x-1\right)>=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}< x< \dfrac{2}{3}\\\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-3\right)\left(5x-1\right)< =0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1}{2}< x< \dfrac{2}{3}\\\dfrac{2}{3}< =x< =3\end{matrix}\right.\)

NV
1 tháng 4 2020

a/ \(\Leftrightarrow x^2-6x+9< 0\)

\(\Leftrightarrow\left(x-3\right)^2< 0\)

BPT vô nghiệm

b/ \(\Leftrightarrow12x^2-3x+1>0\)

\(\Leftrightarrow12\left(x-\frac{1}{8}\right)^2+\frac{13}{16}>0\) (luôn đúng)

Vậy tập nghiệm của BPT là \(D=R\)

c/ \(\Leftrightarrow2\left(x-4\right)\left(x-1\right)\left(x-3\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}1< x< 3\\x>4\end{matrix}\right.\)

6 tháng 4 2020

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7 tháng 4 2020

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NV
8 tháng 3 2020

1. \(\Leftrightarrow\left(2x-1\right)\left(3x+1\right)< 0\)

\(\Rightarrow-\frac{1}{3}< x< \frac{1}{2}\)

2. \(\Leftrightarrow\left(x-2\right)\left(3-2x\right)>0\)

\(\Rightarrow\frac{3}{2}< x< 2\)

3. \(\Leftrightarrow\left(5x-3\right)^2>0\)

\(\Rightarrow x\ne\frac{3}{5}\)

4. \(\Leftrightarrow-3\left(x-\frac{1}{6}\right)-\frac{59}{12}< 0\)

\(\Rightarrow x\in R\)

5. \(\Leftrightarrow2\left(x-1\right)^2+5\ge0\)

\(\Rightarrow x\in R\)

NV
8 tháng 3 2020

6. \(\Leftrightarrow\left(x+2\right)\left(8x+7\right)\le0\)

\(\Rightarrow-2\le x\le-\frac{7}{8}\)

7.

\(\Leftrightarrow\left(x-1\right)^2+2>0\)

\(\Rightarrow x\in R\)

8. \(\Leftrightarrow\left(3x-2\right)\left(2x+1\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge\frac{2}{3}\end{matrix}\right.\)

9. \(\Leftrightarrow\frac{1}{3}\left(x+3\right)\left(x+6\right)< 0\)

\(\Rightarrow-6< x< -3\)

10. \(\Leftrightarrow x^2-6x+9>0\)

\(\Leftrightarrow\left(x-3\right)^2>0\)

\(\Rightarrow x\ne3\)