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Tham khảo:
TH1: Nếu x>2 ta có: x-2 + 2x-3 =2x + 1
<=> x=6 .
TH2: Nếu \(\dfrac{3}{2}\) ≤x≤2 ta có: 2 - x + 2x -3 = 2x+1
<=> x=-2 (loại)
TH3: x<\(\dfrac{3}{2}\) ta có : 2-x+3-2x= 2x+1
<=> x=\(\dfrac{4}{5}\)
Vậy : x=6 ; x=\(\dfrac{4}{5}\)
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Giải
TH1: Nếu x>2 ta có: x-2 + 2x-3 =2x + 1
<=> x=6 .
TH2: Nếu 3/2 ≤x≤2 ta có: 2 - x + 2x -3 = 2x+1
<=> x=-2 (loại)
TH3: x<3/2 ta có : 2-x+3-2x= 2x+1
<=> x=4/5
Vậy : x=6 ; x=4/5
![](https://rs.olm.vn/images/avt/0.png?1311)
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Tìm x biết:
a) 3x-|2x+1|=2
b)2.|5x-3|-2x=14
c)|x+1|+|x+2|+|x+3|=4x
d)|x-2|+|3-2x|=2x+1
e)|x-3|=(-2).|x+4|
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a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)
b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))
\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)
\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)
d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)
\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)
\(\Leftrightarrow2x^2+2x=2x^2+1\)
\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).
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`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`