a) \(-\dfrac{2}{5}+\dfrac{5}{6}x=-\dfrac{4}{15}\\ \Leftrightarrow\dfrac{5}{6}x=\dfrac{2}{15}\\ \Leftrightarrow x=\dfrac{4}{25}\)

b) \(\dfrac{2}{3}+\dfrac{7}{4}\div x=\dfrac{5}{6}\\ \Leftrightarrow\dfrac{7}{4}\div x=\dfrac{1}{6}\\ \Leftrightarrow x=\dfrac{7}{24}\)

a: Ta có: \(-\dfrac{2}{5}+\dfrac{5}{6}x=\dfrac{-4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{5}{6}=\dfrac{2}{15}\)

hay \(x=\dfrac{4}{25}\)

b: Ta có: \(\dfrac{7}{4}:x+\dfrac{2}{3}=\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{7}{4}:x=\dfrac{1}{6}\)

hay \(x=\dfrac{21}{2}\)

a) 6x + 15 = 75

6x = 75 - 15

6x = 60

x = 60 : 6

x = 10

b) x - 12 = (-27)

x = (-27) + 12

x = -15

c) 23 + (x - 5) : 4 = 3⁸ : 3⁵

23 + (x - 5) : 4 = 3³

23 + (x - 5) : 4 = 27

23 + (x - 5) = 27 x 4

23 + (x - 5) = 108

x - 5 = 108 - 23

x - 5 = 85

x = 85 + 5

x = 90

1) 14x-8x=10+5

x(14-8)=15

x6=15

x=15/6

2)5x-3x=30-15

2x=15

x=15/2

3)làm tương tự

1) x=2,5

2) x=7,5

3) x=4

4) x=7/3

5) x=8,25

[ ( 6x - 39 ) : 7 ] . 4 = 12

=> ( 6x - 39 ) : 7 = 12 : 4

=> ( 6x - 39 ) : 7 = 3

=> 6x - 39 = 3.7

=> 6x - 39 = 21

=> 6x = 21+39

=> 6x = 60

=> x = 10.

( x : 3 - 4 ) . 5 = 15

=> x : 3 - 4 = 15 : 5

=> x : 3 - 4 = 3

=> x : 3 = 7

=> x = 7.3

=> x = 21

x - [ 42 + (-28) ] = -8

=> x - 14 = -8

=> x = -8 + 14

=> x = 6.

[ (6x - 39): 7] . 4=12

(6x-39) : 7=12:4

(6x -39) :7=3

(6x-39) =3 . 7

6x-39 =21

6x =21 +39

6x=20

x=10

(x :3 -4) .5 =15

(x : 3 -4) =15:5

x:3-4 =3

x : 3=3+4

x:3 =7

x=7 . 3

x=21

5-x-16=40+x

-x-x=40+16-5

(-x).2=51

-x=51/2=>x=51/2

1.x=40

2.x=10

3.x=

15-x = 4x -5 

15 + 5 = 4x + x 

5x = 20

x= 4

x - 15 = 6 + 4x 

-15 - 6 = 4x -x 

3x = -21 

x= -7

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

\(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)

\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)

\(24x^2+4x-30x-5-24x^2+32x-9x+12=15\)

\(-3x=8\)

\(x=-\frac{8}{3}\)

=.= hok tốt!!

Bạn chỉ áp dụng phương pháp nhân đa thức với đa thức sau 

rút gọn các hạng tử đồng dạng cho nhau thì nó sẽ trở về

dạng tìm x như bình thường

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 

a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21

<=> 12x² + 24x - 6x - 12 + 20x - 12x² - 5 + 3x = -21

<=> 41x = -21 + 12 + 5 

<=> 41x = -4

<=> x = -4/41

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.