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3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Tìm x biết
1. 2(5x-8)-3(4x-5)=4(3x-4)+11
2. (2x+1)2-(4x-1).(x-3)-15=0
3. (3x-1).(2x-7)-(1-3x).(6x-5)=0
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1) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)
\(\Rightarrow17x=17\Rightarrow x=1\)
3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)
\(\Leftrightarrow17x=17\)
hay x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
![](https://rs.olm.vn/images/avt/0.png?1311)
a. x( x+ 3)= 0
⇔ x= 0 hoặc x+ 3= 0
⇔ x= 0 x = -3
b. x( 2x− 1)+ 2( 2x− 1) =0
⇔ ( 2x− 1)(x+ 2) =0
⇔ 2x− 1 =0 hoặc x+ 2 =0
⇔ 2x =1 x = -2
⇔ x =\(\dfrac{1}{2}\) x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)
\(\Leftrightarrow-2x+1-x-2=8\)
\(\Leftrightarrow-3x=9\)
hay x=-3
b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)
\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ( 3x - 1 )^2 - 3x( 3x + 2 ) = 0
<=>9x2-6x+1-9x2-6x=0
<=>-12x+1=0
<=>-12x=-1
<=>x=1/12
b, ( 2x + 3)^2 = 4x(x + 1 )
<=>(2x+3)2-4x(x+1)=0
<=>4x2+12x+9-4x2-4x=0
<=>8x+9=0
<=>8x=-9
<=>x=-9/8
c) vô fx gõ lại
d)x2-4x+4=16
<=>(x-2)2-16=0
<=>(x-2)2-42=0
<=>(x-2+4)(x-2-4)=0
<=>(x+2)(x-6)=0
<=>x+2=0 hoặc x-6=0
<=>x=-2 hoặc x=6
\((4x-1)^2-4(2x-3)^2-x-4=0\\\Leftrightarrow (4x)^2-2\cdot4x\cdot1+1^2-4[(2x)^2-2\cdot2x\cdot3+3^2]-x-4=0\\\Leftrightarrow 16x^2-8x+1-4(4x^2-12x+9)-x-4=0\\\Leftrightarrow 16x^2-8x+1-16x^2+48x-36-x-4=0\\\Leftrightarrow (16x^2-16x^2)+(-8x+48x-x)+(1-36-4)=0\\\Leftrightarrow 39x-39=0\\\Leftrightarrow 39x=39\\\Leftrightarrow x=1\\Vậy:x=1\)
(4x - 1)² -4(2x - 3)² - x - 4 = 0
16x² - 8x + 1 - 4(4x² - 12x + 9) - x - 4 = 0
16x² - 8x + 1 - 16x² + 48x - 36 - x - 4 = 0
39x - 39 = 0
39x = 39
x = 39 : 39
x = 1