(4x-3)-(x+5)=3(10-x)
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\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
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a) Đặt x^2+2x+2=t
\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)
\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)
Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
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\(14x\left(x+3\right)-7x^2\left(3+x\right)=\left(x+3\right)\left(14x-7x^2\right)=7x\left(2-x\right)\left(x+3\right)\\ \left(2x-5\right)\left(3+4x\right)-4x+10=\left(2x-5\right)\left(3+4x-2\right)=\left(2x-5\right)\left(4x+1\right)\)
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Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
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a: \(=2x^2-6x+x-3-20x+8x^2\)
\(=10x^2-25x-3\)
b: \(=x^2+4x+4-2\left(x^2-9\right)+10\)
\(=x^2+4x+14-2x^2+18\)
\(=-x^2+4x+32\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ 2(4x-3)-3(x+5)+4(x-10)= 5(x+2)
-> 8x - 6 - 3x - 15+4x - 40= 5x + 10 (mở ngoạc bạn nhé!)
-> 8x - 6x - 4x - 6 -15 -40 = 5x + 10
-> 9x + 61 = 5x +10
-> 9x - 5x = 10 - 61
-> 4x = -51
-> x = -12.75
Vậy x= -12.75 bạn nhé!
Chúc bạn học tôtr. Nhớ tick cho mk nha!
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( 4x - 3 ) - ( x + 5 ) = 3 . ( 10 - x )
<=> 4x - 3 - x - 5 = 30 - 3x
=> 3x - 8 = 30 - 3x
=> 3x + 3x = 30 + 8
=> 6x = 38
=> x = 19/3
Vậy x = 19/3
(4x−3)−(x+5)=3(10−x)
⇒4x−3−x−5=30−3x
⇒3x−8=30−3x
⇒6x=38
⇒x=\(\dfrac{19}{3}\)
Vậy x= \(\dfrac{19}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
⇒ 8x - 6 - 3x - 15 + 4x - 40 = 5x + 10
⇒ 9x - 61 = 5x + 10
⇒ 4x = 71
⇒ x = 17,7
\(\Rightarrow\) 8x - 6 - 3x - 15 + 4x - 40 = 5x + 10
\(\Rightarrow\) 9x - 61 = 5x + 10
\(\Rightarrow\) 4x = 71
\(\Rightarrow\) x = 17,75
![](https://rs.olm.vn/images/avt/0.png?1311)
e) 5(3-2x)+5(x-4)=6-4x
⇔15-10x+5x-20-6+4x=0
⇔-x-11=0
hay -x=11
⇔x=-11
Vậy: x=-11
g) 2(4x-8)-7(3+x)=|-4|(3-2)
⇔8x-16-21-7x=4
⇔x-37=4
hay x=41
Vậy: x=41
k) -7(5-x)-2(x-10)=15
⇔-35+7x-2x+20=15
⇔5x-15=15
hay 5x=30
⇔x=6
Vậy: x=6
m) -4(x+1)+(89x-3)=24
⇔-4x-4+89x-3=24
⇔85x-7=24
⇔85x=31
hay \(x=\frac{31}{85}\)(ktm x∈Z)
Vậy: x∈∅
q) 10(x-7)-8(x+5)=6*(-5)+24
⇔10x-70-8x-40=-6
⇔2x-110=-6
⇔2x=104
hay x=52
Vậy: x=52
4x-3-x-5=30-3x
3x-3-5=30-3x
3x-8=30-3x
3x+3x=30+8
6x=38
x=38:6
x=38/6
(4x-3)-(x+5)=3(10-x)
<=>4x-3-x-5=30-3x
<=>4x-3-x-5-30+3x=0
<=>6x-38=0
<=>6x=38
<=>x=\(\frac{19}{3}\)