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(3x - 2)(2x + 3) - (6x2 - 85) - 99 = 0
(3x - 2)(2x + 3) - 6x2 + 85 - 99 = 0
(3x - 2)(2x + 3) - 6x2 - 14 = 0
6x2 + 9x - 4x - 6 - 6x2 - 14 = 0
5x - 20 = 0
5x = 0 + 20
5x = 20
x = 20 : 5
x = 5
=> x = 5
2x + 2{-[-x + 3(x - 3)]} = 2
2x + 2[x - 3(x - 2)] = 2
2x + 2x - 6x + 18 = 2
-2x + 18 = 2
-2x = 2 - 18
-2x = -16
x = (-16) : (-2)
x = 8
=> x = 8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(6x^2-2x\left(3x+\dfrac{3}{2}\right)=9\)
\(\Rightarrow6x^2-6x^2-3x=9\)
\(\Rightarrow-3x=9\)
\(\Rightarrow x=\dfrac{9}{-3}\)
\(\Rightarrow x=-3\)
\(6x^2-2x\left(3x+\dfrac{3}{2}\right)=9\\ \Leftrightarrow6x^2-6x^2-3x=9\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
15 \(\times\) ( 2\(x\) - 16) - (6\(x^2\) + 15\(x\)): 3\(x\) = 20
15 \(\times\) (2\(x\) - 16) - 3\(x\)( 2\(x\) + 5):3\(x\) = 20
30\(x\) - 240 - (2\(x\) + 5) = 20
30\(x\) - 240 - 2\(x\) - 5 = 20
28\(x\) - 245 = 20
28\(x\) = 20 + 245
28\(x\) = 265
\(x\) = 265:28
15(2x-16)-(6\(x^2\)+15x):3x=20
=>30x-240-2x-5=20
=>28x=265
=>x=\(\dfrac{265}{28}\)
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+) \(2x\left(x-4\right)-x\left(2x+3\right)+22=0\)
\(\Leftrightarrow2x^2-8x-2x^2-3x+22=0\)
\(\Leftrightarrow-11x+22=0\)
\(\Leftrightarrow-11\left(x-2\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
+) \(\left(2x+3\right)\left(3x+2\right)+2\left(1-3x\right)\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow6x^2+4x+9x+6+\left(2-6x\right)\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow6x^2+13x+6+2x+1-6x^2-3x=1\)
\(\Leftrightarrow12x+7=1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
2x( x - 4 ) - x( 2x + 3 ) + 22 = 0
<=> 2x2 - 8x - 2x2 - 3x + 22 = 0
<=> -11x + 22 = 0
<=> -11x = -22
<=> x = 2
( 2x + 3 )( 3x + 2 ) + 2( 1 - 3x )( x + 1/2 ) = 1
<=> 6x2 + 13x + 6 + 2( -3x2 - 1/2x + 1/2 ) = 1
<=> 6x2 + 13x + 6 - 6x2 - x + 1 = 1
<=> 12x + 7 = 1
<=> 12x = -6
<=> x = -6/12 = -1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
(2\(x\) - 1).(2\(x\) - 5) < 0
Lập bảng ta có:
\(x\) | \(\dfrac{1}{2}\) \(\dfrac{5}{2}\) |
2\(x\) - 1 | - 0 + + |
2\(x\) - 5 | - - 0 + |
(2\(x\) - 1).(2\(x\) - 5) | + 0 - 0 + |
Theo bảng trên ta có: \(\dfrac{1}{2}\) < \(x\) < \(\dfrac{5}{2}\)
(3 - 2\(x\)).(\(x\) + 2) > 0
Lập bảng ta có:
\(x\) | -2 \(\dfrac{3}{2}\) |
3 - 2\(x\) | + + 0 - |
\(x\) + 2 | - 0 + + |
(3 -2\(x\)).(\(x\) +2) | - 0 + 0 - |
Theo bảng trên ta có: - 2 < \(x\) < \(\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(6x^2+2x+2=0\)
\(\text{Δ}=2^2-4\cdot6\cdot2=4-48=-44< 0\)
Do đó: Phương trình vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
`#040911`
`a)`
`2x^2 - 3x = 0`
`\Rightarrow x(2x - 3) = 0`
`\Rightarrow`\(\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=0\\2x=3\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy, \(x\in\left\{0;\dfrac{3}{2}\right\}\)
`b)`
\(x+\dfrac{1}{2}-z-\dfrac{2}{3}=\dfrac{1}{2}?\)
Bạn xem lại đề
`c)`
\(x^3-x^2=0\\ \Rightarrow x^2\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy, \(x\in\left\{0;1\right\}.\)
\(a,2x^2-3x=0\\ \Leftrightarrow x\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\\ b,Xem.lại,đề\\ c,x^3-x^2=0\\ \Leftrightarrow x^2.\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)