a) \(5\left(x-7\right)=0\)

\(\Rightarrow x-7=0\)

\(\Rightarrow x=7\)

b) \(25\left(x-4\right)=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

c) \(\left(34-2x\right)\left(2x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)

d) \(\left(2019-x\right)\left(3x-12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)

e) \(57\left(9x-27\right)=0\)

\(\Rightarrow9x-27=0\)

\(\Rightarrow9\left(x-3\right)=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

a) 5.(x-7)=0⇔x-7=0⇔x=7

b) 25(x-4)=0⇔x-4=0⇔x=4

c) (34-2x).(2x-6)=0

⇔ 34-2x=0 hoặc 2x-6=0

⇔2x=34 hoặc 2x=6

⇔ x=17 hoặc x=3

d) (2019-x).(3x-12)=0

⇔ 2019-x=0 hoặc 3x-12=0

⇔ x=2019 hoặc x=4

e) 57.(9x-27)=0

⇔ 9x-27=0

⇔ x=3

f) 25+(15-x)=30

⇔ 15-x=5

⇔ x=10

g) 43-(24-x)=20

⇔ 24-x=23

⇔ x=1

h) 2.(x-5)-17=25

⇔ 2(x-5)=42

⇔x-5=21

⇔ x=26

i) 3(x+7)-15=27

⇔ 3(x+7)=42

⇔ x+7=14

⇔ x=7

j) 15+4(x-2)=95

⇔ 4(x-2)=80

⇔ x-2=20

⇔ x=22

k) 20-(x+14)=5

⇔ x+14=15

⇔ x=1

l) 14+3(5-x)=27

⇔ 3(5-x)=13

⇔ 5-x=13/3

⇔ x=5-13/3

⇔ x=2/3

`@` `\text {Ans}`

`\downarrow`

`c)`

`( 34 - 2x ) . ( 2x - 6 ) = 0`

`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)

Vậy, `x \in {17; 3}`

`d)`

`( 2019 - x ) . ( 3x - 12 ) =0` `?`

`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)

Vậy, `x \in {2019; 4}`

`e) `

`57 . ( 9x - 27 ) = 0`

`=>`\(9x-27=0\div57\)

`=> 9x - 27 = 0`

`=> 9x = 27`

`=> x = 27 \div 9`

`=> x = 3`

Vậy, `x = 3`

`f)`

`25 + ( 15 - x ) = 30`

`=> 15 - x = 30 - 25`

`=> 15 - x = 5`

`=> x = 15 -5 `

`=> x = 10`

Vậy, `x = 10`

`g) `

`43 - ( 24 - x ) = 20`

`=> 24 - x = 43 - 20`

`=> 24 - x = 23`

`=> x = 24 - 23`

`=> x = 1`

Vậy, `x = 1`

`h) `

`2 . ( x - 5 ) - 17 = 25`

`=> 2 ( x - 5) = 25+17`

`=> 2 ( x - 5) = 42`

`=> x - 5 = 42 \div 2`

`=> x - 5 = 21`

`=> x = 21 + 5`

`=> x = 26`

Vậy, `x = 26`

`i)`

`3 . ( x + 7 ) - 15 = 27`

`=> 3(x + 7) = 27 + 15`

`=> 3(x + 7) = 42`

`=> x +7 = 42 \div 3`

`=> x + 7 = 14`

`=> x = 14 - 7`

`=> x = 7`

Vậy, `x = 7`

`j)`

`15 + 4 . ( x - 2 ) = 95`

`=> 4(x - 2) = 95 - 15`

`=> 4(x - 2) = 80`

`=> x - 2 = 80 \div 4`

`=> x - 2 = 20`

`=> x = 20 + 2`

`=> x = 22`

Vậy, `x = 22`

`k)`

`20 - ( x + 14 ) = 5`

`=> x + 14 = 20 - 5`

`=> x + 14 = 15`

`=> x = 15 - 14`

`=> x = 1`

Vậy, `x = 1`

`l) `

`14 + 3 . ( 5 - x ) = 27`

`=> 3(5 - x) = 27 - 14`

`=> 3(5 - x) = 13`

`=> 5 - x = 13 \div 3`

`=> 5 - x = 13/3`

`=> x = 5- 13/3`

`=> x = 2/3`

Vậy, `x = 2/3.`

`@` `\text {Kaizuu lv uuu}`

nhanh mik tick cho nha

a/x=0

b/90

c/121

d/15

e/15

a) 0

b)  x - 15 =75

x=15+75=

x=90

c)  ( x - 7 ) = 114+0

      x - 7 = 114

x=114+7

x=121

d)  ( x - 15 ) = 0:2015

       x - 15 =0

  x=15+0

x=15

e)( x - 15 ) =0:13

x - 15 =0

x=15+0

x+15

a. 25 . x = x

x = 0

b. ( x - 15 ) - 75 = 0

x - 15 = 0 + 75 

x - 15 = 75

x = 75 + 15

x = 90

c. 114 - ( x - 7 ) = 0

x - 7 = 114 - 0

x - 7 = 114

x = 114 + 7

x = 121

d. 2015 x ( x - 15 ) = 0

x - 15 = 0 : 2015

x - 15 = 0

x = 0 + 15

x = 15

e. ( x - 15 ) x 13 = 0

x - 15 = 0 : 14

x - 15 = 0

x = 0 + 15

x = 15 

Tương tự 3. HS tự làm.

Yêu cầu đề là gì vậy bạn? Nếu là tìm y thì phải có dấu "=" chứ nhỉ?

ủa lớp 5 mà có bài này hả
có thiếu đề ko em

em viết thiếu ah

a) \(\left(x-2018\right)\cdot3=0\)

\(\Leftrightarrow x-2018=0\)

\(\Leftrightarrow x=2018\)

b) \(25\left(x-8\right)=0\)

\(\Leftrightarrow x-8=0\)

\(\Leftrightarrow x=8\)

c) \(25+\left(15+x\right)=75\)

\(\Leftrightarrow15+x=50\)

\(\Leftrightarrow x=35\)

a) x=2018

b)x=8

c)x=35

Bài 2:

a, (x-2018).3=0

<=> x-2018=0

<=>x=0+2018

<=>x=2018

Vậy x=2018

b, 25(x-8)=0

<=> x-8=0

<=> x=0+8

<=>x=8

Vậy x=8

c, 25+(15+x)=75

<=> 25+15+x=75

<=> 40+x=75

<=>x=75-40

<=>x=35

Vậy x=35

`@` `\text {Ans}`

`\downarrow`

`25 + (15 - x) = 0`

`25 + 15 - x = 0`

`40 - x = 0`

`x = 40 - 0`

`x = 40.`

Vậy, `x = 40.`

=>40-x=0

=>x=40-0

=>x=40