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\(\dfrac{2x-3}{3}=\dfrac{27}{2x-3}\)
=>(2x-3)2=27.3
2x-3=9 hoặc 2x-3= -9
x=6 hoặc x=-3
\(\dfrac{2x-3}{3}=\dfrac{27}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ \Leftrightarrow\left(2x-3\right)^2=27.3=81\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=9\\2x-3=-9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\left(tmđk\right)\\x=-3\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x\in\left\{6;-3\right\}\)
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a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 )
<=> 2x - 3 - x + 5 = x + 7 - x - 2
<=> x = 3
b)(7x-5)-(6x+4)=(2x+3)-(2x+1)
<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1
<=> x = 11
c)(9x-3)-(8x+5)=(3x+2)
<=> 9x - 3 - 8x - 5 = 3x + 2
<=> -2x = 10
<=> x = -5
d)(x+7)-(2x+3)=(3x+5)-(2x+4)
<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4
<=> -2x = -3
<=> x = 3/2
ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠ
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a) 3x - 4 (2x-7) = 3x - 3
3x - 8x + 28 = 3x - 3
=> 3x - 8x - 3x = -3 -28
-8x = -31
x = 31/8
b) 3(2x-7) - (4-2x) = 2x -1
6x - 7 - 4 + 2x = 2x -1
=> 6x + 2x - 2x = -1 + 7+4
6x = 10
x = 5/3
c) (x-2).(2x+12)=0
=> x -2 = 0 => x = 2
2x + 12 = 0 => 2x = -12 => x = -6
KL: x = 2 hoặc x = -6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x + 1) + (x - 3) = x - 2
=> x + 1 + x - 3 - x + 2 = 0
=> x = 0
b) - (1 - x) - (2x - 3) = 4 - 2x
=> -1 + x - 2x + 3 - 4 + 2x = 0
=> x - 2 = 0
=> x = 2
c)-13 - (5 - x + 6) + 2x = 3
=> -13 - 5 + x - 6 + 2x - 3 = 0
=> 3x - 27 = 0
=> 3x = 27
=> x = 9
![](https://rs.olm.vn/images/avt/0.png?1311)
3) |x| = 3+2x
Th1: x= 3 +2x
x -2x = 3
-x = 3
x= -3
Th2: x= -(3+2x)
x= -3 -2x
x+2x = -3
3x = -3
x= -1
Vậy x= -3; x=-1
4) | 3-2x| =4/3
Th1: 3-2x = 4/3
2x = 3- 4/3
2x = 5/3
x= 5/3 : 2
x= 5/6
Th2: 3 - 2x = -4/3
2x = 3- (-4/3)
2x = 13/3
x = 13/3 : 2
x = 13/6
Vậy x= 5/6 và x= 13/6
3)Ta có: |x| = 3+2x
<=> x = 3 + 2x
(-x) = 3 + 2x
<=> 3 = 2x - x
3 = 2x - (-x)
<=> x = 3
3x = 3
<=> x = 3
x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a, => 3^2x = 3^4
=>2x=4
=>x=2
b,=>2x+3x-3=17
=>5x-3=17
=>5x=17+3=20
=>x=4
c,=>x+15=(2x+3).2 = 4x+6
=>4x+6-(x+15)=0
=>4x+6-x-15=0
=>3x-9=0
=>3x=9
=>x=3
k mk nha
\(\left(2x-3\right)^3-\left(2x-3\right)=0\)
\(\left(2x-3\right)\left[\left(2x-3\right)^2-1\right]=0\)
\(\left(2x-3\right)\left(2x-4\right)\left(2x-2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=2\\x=1\end{cases}}\)
Vì (2x-3)3=2x-3 =>2x-3=1 hoặc 2x-3=-1
*2x-3=1 *2x-3=-1
2x=1+3 2x=-1+3
2x=4 2x=2
x=4:2 x=2:2
x=2 x=1
Vậy x thuộc {1;2}