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Bài 1:
c) |2x - 1| = x + 2
<=> 2x - 1 = +(x + 2) hoặc -(x + 2)
* 2x - 1 = x + 2
<=> 2x - x = 2 + 1
<=> x = 3
* 2x - 1 = -(x + 2)
<=> 2x - 1 = x - 2
<=> 2x - x = -2 + 1
<=> x = -1
Vậy.....
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4\left(x-3\right)^2-\left(2x-1\right)^2>12x\)
\(\Leftrightarrow\)\(4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)>12x\)
\(\Leftrightarrow\)\(4x^2-24x+36-4x^2+4x-1>12x\)
\(\Leftrightarrow\)\(-20x+35-12x>0\)
\(\Leftrightarrow\)\(-32x+35>0\)
\(\Leftrightarrow\)\(-32x>-35\)
\(\Leftrightarrow\)\(x< \frac{35}{32}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(4x-1\right)\left(x^2+12\right)\left(-x+4\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1>0\\x^2+12>0\left(LD\forall x\right)\\-x+4>0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x>1\\-x>-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{4}\\x< 4\end{matrix}\right.\)
Vậy \(S=\left\{x|\dfrac{1}{4}< x< 4\right\}\)
\(b,\left(2x-1\right)\left(5-2x\right)\left(1-x\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1< 0\\5-2x< 0\\1-x< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\)
Vậy \(S=\left\{x|1>x>\dfrac{5}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< \dfrac{x^2}{7}-\dfrac{2x-3}{5}\)
\(\Leftrightarrow2x-3+5x\left(x-2\right)< 5x^2-7\left(2x-3\right)\)
\(\Leftrightarrow2x-3+5x^2-10x< 5x^2-14x+21\)
=>-8x-3<-14x+21
=>6x<24
hay x<4
3: \(\dfrac{3x-2}{4}< \dfrac{3x+3}{6}\)
\(\Leftrightarrow3\left(3x-2\right)< 2\left(3x+3\right)\)
=>9x-6<6x+6
=>3x<12
hay x<4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2x-3}{35}\) + \(\dfrac{x\left(x-2\right)}{7}\) < \(\dfrac{x^2}{7}\) - \(\dfrac{2x-3}{5}\)
<=> \(\dfrac{2x-3}{35}\) + \(\dfrac{5x\left(x-2\right)}{7.5}\) < \(\dfrac{5x^2}{7.5}\) - \(\dfrac{7\left(2x-3\right)}{7.5}\)
<=> 2x-3 + 5x2-10x < 5x2 - 14x + 21
<=> 5x2 - 5x2 + 2x -10x + 14x < 21 + 3
<=> 6x < 24
<=> x < 4
vậy bpt có tập nghiệm S={ x < 4 }
![](https://rs.olm.vn/images/avt/0.png?1311)
Th1
2x+3=x-4(x>=-3/2)
<=>x=-7(loại)
Th2
2x+3=4-x(x=<-3/2)
<=>3x=1
<=>x=1/3(loại)
Pt vô nghiệm
\(\left|2x+3\right|=x-4\left(x\ge4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=x-4\\2x+3=4-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3-4\\3x=4-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\left(L\right)\\x=\dfrac{1}{3}\left(L\right)\end{matrix}\right.\)
Không có giá trị của x thỏa mãn.