a ) \(5.\left(x-7\right)=0\)

\(\Leftrightarrow x-7=5.0\)

\(\Leftrightarrow x-7=0\)

\(\Leftrightarrow x=7\)

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

\(\Leftrightarrow2.x-6=34.0\)

\(\Leftrightarrow2.x-6=0\)

\(\Leftrightarrow2.x=0+6\)

\(\Leftrightarrow2.x=6\)

\(\Leftrightarrow x=6\div2\)

\(\Leftrightarrow x=3\)

c ) \(25+\left(15-x\right)=30\)

\(\Leftrightarrow25+\left(15-x\right)=-\left(x-40\right)\)

\(\Leftrightarrow-\left(x-40\right)=2.3.5\)

\(\Leftrightarrow40-x=30\)

\(\Leftrightarrow-x=-10\)

\(\Leftrightarrow x=10\)

d ) \(43-\left(24-x\right)=20\)

\(\Leftrightarrow24-x=43-20\)

\(\Leftrightarrow24-x=23\)

\(\Leftrightarrow x=24-23\)

\(\Leftrightarrow x=1\)

e ) \(2.\left(x-5\right)-17=25\)

\(\Leftrightarrow2.\left(x-5\right)=25+17\)

\(\Leftrightarrow2.\left(x-5\right)=42\)

\(\Leftrightarrow x-5=42\div2\)

\(\Leftrightarrow x-5=21\)

\(\Leftrightarrow x=21+5\)

\(\Leftrightarrow x=26\)

f ) \(24+3.\left(5-x\right)=27\)

\(\Leftrightarrow3.\left(5-x\right)=27-24\)'

\(\Leftrightarrow3.\left(5-x\right)=3\)

\(\Leftrightarrow5-x=3\div3\)

\(\Leftrightarrow5-x=1\)

\(\Leftrightarrow x=5-1\)

\(\Leftrightarrow x=4\)

g ) \(15\div x-2=3\)

\(\Leftrightarrow15\div x=3+2\)

\(\Leftrightarrow15\div x=5\)

\(\Leftrightarrow x=15\div5\)

\(\Leftrightarrow x=3\)

h ) \(\left(32-x\div5\right)\div13=2\)

\(\Leftrightarrow32-x\div5=2\times13\)

\(\Leftrightarrow32-x\div5=26\)

\(\Leftrightarrow x\div5=32-26\)

\(\Leftrightarrow x\div5=6\)

\(\Leftrightarrow x=5.6\)

\(\Leftrightarrow x=30\)

g ) \(\left(6-2x\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}6-2x=0\\x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

Vậy \(x\in\left\{3;8\right\}\)

Mình làm hết nhaa !!

a)5.(x-7)=0

x-7=0:5

x-7=0

x=0+7

x=7

Vậy x=7

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

2x-6 =0:34

2x-6 =0

2x=0+6

2x=6

x=6:2

x=3

Vậy x= 3

c)25+(15-x)=30

15-x=30-25

15-x=5

x=15-5

x=10

Vậy x=10

d)43-(24-x)=20

24-x =43-20

24-x=23

x=24-23

x=1

vậy x=1

e)2(x-5)-17=25

2(x-5)=25+17

2(x-5)=42

x-5=42:2

x-5=21

x=21+5

x=26

Vậy x=26

f)24+3(5-x)=27

3(5-x)=27-24

3(5-x)=3

5-x =3:3

5-x=1

x=5-1

x=4

Vậy x=4

g)15:x-2=3

15:x=3+2

15:x=5

x=15:5

x=3

Vậy x=3

h)( 32-x:5):13=2

32-x:5 =2.13

32-x :5=26

x:5= 32-26

x:5=6

x=6.5

x=30

Vậy x=30

k)(6-2x)(x-8)=0

➩6-2x=0 hoặc x-8=0

2x=6-0 x=0+8

2x=6 x=8

x=6:2

x=3

Vậy x=3 hoặc x=8

a) 25. (x-4) = 0

=> x -4 =0 

x = 4

b) 43 - (24-x) = 20

43 - 24 + x = 20

19 + x = 20

x = 1

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

3.x + 21 - 15 = 27

3.x + 6 = 27

3.x = 21

x = 7

d)... bn ghi thiếu đề r

e) (2.x-6).(x-7) = 0

=> 2.x -6 = 0 => 2x = 6 => x = 3

x - 7 = 0 => x  = 7

KL: x = 3 hoặc x = 7

phần d lm tương tự như phần f nha bn!
 

a,25(x-4)=0

x-4=0

x=4

b,43-(24-x)=20

43-24+x=20

x=1

c,3(x+7)-15=27

3x+21-15=27

3x=21

x=7

d,(x-4)(x-12)=0

x-4=0=>x=4

x-12=0=>x=12

e,(2x-6)(x-7)=0

2x-6=0=>x=3

x-7=0=>x=7

f,(5x-10)(2x-8)=0

5x-10=0=>x=2

2x-8=0=>x=2

a) 25 ( x - 4 ) = 0

<=>     x - 4   = 0

<=>      x       = 4

Vậy x = 4

b) 43 - ( 24 - x ) = 20 

<=>       24 - x   = 43 - 20

<=>       24 - x   = 23

<=>              x   = 1

Vậy x = 1

c) 3( x + 7 ) - 15 = 27 

<=> 3 ( x + 7 )    = 27 + 15

<=> 3 ( x + 7 )    = 42

<=>       x + 7     = 42 : 3

<=>       x + 7     = 14

<=>        x          = 7

Vậy x = 7

d)

e) ( 2x - 6 ) ( x - 7 ) = 0

\(\Leftrightarrow\) \(\orbr{\begin{cases}2x-6=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=6\\x=7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}\)

Vậy x = 3 ; x = 7

f) ( 5x - 10 ) ( 2x - 8 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}5x-10=0\\2x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}5x=10\\2x=8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)

Vậy x = 2 ; x = 4 

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,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

`@` `\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

1. Tìm x biết:
a) 15 – ( 4 – x) = 6

4-x=15-6

4-x=9

x=4-9

x=-5

b) - 30 + ( 25 – x) = - 1

25-x=-1-(-30)

25-x=29

x=25-29

x=-4
c) x – ( 12 – 25) = -8

x+13=-8

x=-8-13

x=-21

d) ( x – 29 ) – ( 17 – 38 ) = - 9

(x-29)-(-21)=-9

(x-29)=-9-21

x-29=-30

x=-30+29

x=-1
2. Tìm số nguyên x biết:
a) x – 5 = - 1

x=-1+5

x=4

b) x + 30 = - 4

x=-4-30

x=-34
c) x – ( - 24) = 3

x+24=3

x=3-24

x=-21

d) 22 – ( - x ) = 12

22+x=12

x=12-22

x=-10
e) ( x + 5 ) + ( x – 9 ) = x + 2

x+5+x-9=x+2

x+x-x=2-5+9

x=6

f) ( 27 – x ) + ( 15 + x ) = x – 24

27-x+15+x=x-24

-x+x+x=-24-27-15

x=-66
3. Tìm x biết:
a) 461 + (x - 45) = 387

x-45=387-461

x-45=-74

x=-74+45

x=-29

b) 11 - (-53+ x ) = 97

-53+x=11-97

-53+x=-86

x=-86-(-53)

x=-33

c) - ( x + 84) + 213 = -16

-(x+84)=-16-213

-(x+84)=-229

x+84=229

x=229-84

x=145

CÁC BẠN GIÚP MIK CÂU NÀO CX ĐC HẾT, GIẢI MIK TICK NHÉ

Do \(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)-3}{x-2}=5\Rightarrow\) chọn \(f\left(x\right)=5\left(x-2\right)+3=5x-7\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-7+6}-\sqrt[3]{x+25}}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-1}-3+3-\sqrt[3]{x+25}}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{5\left(x-2\right)}{\sqrt[]{5x-1}+3}-\dfrac{x-2}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\left(\dfrac{5}{\sqrt[]{5x-1}+3}-\dfrac{1}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}\right)=\dfrac{5}{3+3}-\dfrac{1}{9+9+9}=\dfrac{43}{54}\)