Ta có: \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}-\dfrac{3}{7}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{7}{6}\end{matrix}\right.\)

Bài 1:

a, 2x + 35 = x - 27

    2x - x = -27 - 35

    x = -62

 b, 2x - 41 = 3x + (-23)

     2x - 3x = -23 + 41

     -x = 18

     x = -18

c, 4 . (x - 3) - 3 . (x - 5) = 45 . (-2) - 34

   4x - 12 - 3x + 15 = -90 - 34

   4x - 3x = -90 - 34 + 12 - 15

   x = -127

Bài 2:

- ( -1234 + 345 - 29 ) - ( 1234 + 135 - 59 )

= 1234 - 345 + 29 - 1234 - 135 + 59

= (1234 - 1234) - (345 + 135) + (29 + 59)

= 0 - 480 + 88

= -392

a)2x+35=x-27

2x-x=-27-35

1x=-62

x=-62:1

X=-62

1, a, 2x + 35 = x - 27 

<=> 2x - x = -27 - 35 

,<=> x = -72 

Vậy x = -72

b, 2x - 41 = 3x + ( - 23 )

<=> 2x - 3x = 41 - 23

<=> -1x = 18

<=> x = -18 

Vậy x = -18 

c, tương tự dùng phân phối .

2, = 1234 - 345 + 29 - 1234 - 135 + 59 

= ( 1234 - 1234 ) + ( -345 - 135 ) + ( 29 + 59 )

= 0 - 480 + 88 = -480 + 88 = -392

a) Ta có: f(x)=-3

<=>x⁵-2x²+x⁴-x⁵+3x²-x⁴-3+2x=-3

<=>(x⁵-x⁵)+(-2x²+3x²)+(x⁴-x⁴)+2x-3=-3

<=>x²+2x-3=-3

<=>x²+2x=0

<=>x(x+2)=0

<=>x=0 hoặc x+2=0

<=>x=0 hoặc x=-2

Vậy..........

b)đa thức f(x) có nghiệm

<=>f(x)=0

<=>x²+2x-3=0

<=>x²+3x-x-3=0

<=>x(x+3)-(x+3)=0

<=>(x-1)(x+3)=0

<=>x-1=0 hoặc x+3=0

<=>x=1 hoặc x=-3

Vậy nghiệm của đa thức f(x) là x=-3;x=1

a, f(x)=0

b, x=1 và x=-3

a: \(A=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{12x^2}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{-x^2-6x-9+x^2-6x+9-12x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-\left(x+1\right)}{x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-12x^2-12x}\)

\(=\dfrac{-\left(x+1\right)\cdot\left(x+3\right)}{-12x^2\left(x+1\right)}=\dfrac{x+3}{12x^2}\)

b: Ta có: |2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>x=-2

Thay x=-2 vào A, ta được:

\(A=\dfrac{-2+3}{12\cdot\left(-2\right)^2}=\dfrac{1}{48}\)

c: Để \(A=\dfrac{2x+1}{x^2}\) thì \(\dfrac{x+3}{12x^2}=\dfrac{2x+1}{x^2}\)

=>x+3=24x+12

=>24x+12=x+3

=>23x=-9

hay x=-9/23

d: Để A<0 thì x+3<0

hay x<-3