tìm x biết:

(3x-1) [- 1/2x+5]=0

1/4+1/3:(2x-1)=-5

[2x+3/5]² - 9/25=0

-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6

[x+1/2]x [2/3-2x]=0

17/2-|2x-3/4|=-7/4

2/3x-1/2x =5/12

(x+1/5)²+17/25=26/25

[x.44/7+3/7].11/5-3/7=-2

3[3x-1/2]+1/9=0

Toán lớp 6Tìm x

 Trả lời  Câu hỏi tương tự

Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !

3(3x-1/2)^3+1/9=0

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

a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)

Vậy: \(S=\left\{-1\right\}\)

b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)

Vậy: \(S=\left\{15\right\}\)

c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)

Vậy: \(S=\left\{5\right\}\)

d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)

Vậy: \(S=\left\{12\right\}\)

e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

a)3x-2=2x-3

=>x=-1

b)7-2x=22-3x

=>x=15

c)8x-3=5x+12

=>3x=15

=>x=5

d)x-12+4x=25+2x-1

=>3x=12

=>x=4

e)x+2x+3x-19=3x+5

=>3x=24

=>x=8

a)3x-2=2x-3

=>x=-1

b)7-2x=22-3x

=>x=15

c)8x-3=5x+12

=>3x=15

=>x=5

d)x-12+4x=25+2x-1

=>3x=36

=>x=12

e)x+2x+3x-19=3x+5

=>3x=24

=>x=8

a. x- (17 - x) = x-7

x-17+x = x+7

2x-17=x+7

2x-x= 7+17

x= 24

b. 9-25=(7-x)-(25+7)

-16=7-x-25-7

-16=-x-25

x= -25 + 16

x = -9

c. (-12)^2 .x = 56 + 10.13.x

144x = 56 + 130x

144x - 130x = 56

14x=56

x= 56 : 14

x= 4

d. 4/24 : (3x-2) = -3

4/24 = -3.(3x-2)

4/24 = -9x + 6

9x = 6 - 4/24

9x = 35/6

x= 35/6 :9

x= 35/54

e. 5/-45 chia hay nhân với (-3-2x) nhỉ

chia bạn nhé

`@` `\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

Gọi a,b,c,... cho dễ nhé!

a,\(7+2x=22-3x\)

\(\Leftrightarrow2x+3x=22-7\)

\(\Leftrightarrow5x=15\)

\(\Leftrightarrow x=3\)

Vậy...

b,\(x-12+4x=25+2x-1\)

\(\Leftrightarrow x+4x-2x=25-1+12\)

\(\Leftrightarrow3x=36\)

\(\Leftrightarrow x=12\)

Vậy...

c,\(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow-2x+x=-4+4-7\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\)

Vậy...

d,\(8x-3=5x+12\)

\(\Leftrightarrow8x-5x=12+3\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=5\)

Vậy...

e,\(x+2x+3x-19=3x+5\)

\(\Leftrightarrow x+2x+3x-3x=5+19\)

\(\Leftrightarrow3x=24\)

\(\Leftrightarrow x=8\)

Vậy...

f,\(\left(x-1\right)-\left(2x-1\right)=9-x\)

\(\Leftrightarrow x-1-2x+1=9-x\)

\(\Leftrightarrow x-2x+x=9-1+1\)

\(\Leftrightarrow0x=9\) (Vô lý)

Vậy...

a, \(7+2x=22-3x\)

\(\Rightarrow7+2x-22+3x=0\)

\(\Rightarrow5x-15=0\)

\(\Rightarrow5x=15\Rightarrow x=3\)

b, \(x-12+4x=25+2x-1\)

\(\Rightarrow3x-12-24-2x=0\)

\(\Rightarrow x-36=0\Rightarrow x=36\)

c, \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Rightarrow7-2x-4=-x-4\)

\(\Rightarrow3-2x+x+4=0\)

\(\Rightarrow-x=-7\Rightarrow x=7\)

d, \(8x-3=5x+12\)

\(\Rightarrow8x-3-5x-12=0\)

\(\Rightarrow3x-15=0\)

\(\Rightarrow3x=15\Rightarrow x=5\)

e, \(x+2x+3x-19=3x+5\)

\(\Rightarrow6x-19-3x-5=0\)

\(\Rightarrow3x-24=0\)

\(\Rightarrow3x=24\Rightarrow x=8\)

f, \(\left(x-1\right)-\left(2x-1\right)=9-x\)

\(\Rightarrow x-1-2x+1-9+x=0\)

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