\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)

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\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

\(14x\left(x+3\right)-7x^2\left(3+x\right)=\left(x+3\right)\left(14x-7x^2\right)=7x\left(2-x\right)\left(x+3\right)\\ \left(2x-5\right)\left(3+4x\right)-4x+10=\left(2x-5\right)\left(3+4x-2\right)=\left(2x-5\right)\left(4x+1\right)\)

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 ....

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

x=-15/6

a: \(=2x^2-6x+x-3-20x+8x^2\)

\(=10x^2-25x-3\)

b: \(=x^2+4x+4-2\left(x^2-9\right)+10\)

\(=x^2+4x+14-2x^2+18\)

\(=-x^2+4x+32\)

tìm x với x \(\in\) R à ..

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!

\(2\left(4x-3\right)-3\left(x+5\right)+4\left(x-10\right)=5\left(x+2\right)\)

\(8x-6-3x-15+4x-40=5x+10\)

( 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

x= \(\dfrac{19}{3}\)

⇒ 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

⇒ 8x - 6 - 3x - 15 + 4x - 40 = 5x + 10

⇒ 9x - 61 = 5x + 10

⇒ 4x = 71

⇒ x = 17,7

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

1 con người làm công việc nghàn đô với bộ óc lớp 6 nể phục