1)120 + ( 34 + 3x ) - ( 6 + x ) = 0

=> 120 + 34 + 3x - 6 - x = 0

=> ( 120 + 34 - 6 ) + ( 3x - x ) = 0

=> 148 + 2x = 0

=> - 2x = 148 => x = 148 : ( - 2 )

=> x = -74

Vậy x = - 74

2)\(\dfrac{-5}{6}-x=\dfrac{7}{12}+\left(\dfrac{-1}{3}\right)\)

\(=>x=\dfrac{-5}{6}-\left(\dfrac{7}{12}-\dfrac{1}{3}\right)\)

\(=>x=\dfrac{-5}{6}-\dfrac{7}{12}+\dfrac{1}{3}\)

\(=>x=\dfrac{-10}{12}-\dfrac{7}{12}+\dfrac{4}{12}\)

\(=>x=\dfrac{-10-7+4}{12}=\dfrac{-13}{12}\)

Vậy x = \(\dfrac{-13}{12}\)

tick cho mk nha

nhớ k đấy mình nhanh nhất

mình nhầm

(3x + 34) - ( 7 + x ) = 186 - 120

3x + 34 - 7 - x        = 66

2x + 27                  = 66

2x                          = 66 - 27

2x                          = 39

x                            = 39 : 2             

x                            = 19,5

`@` `\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}.`

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

`@` `\text {Ans}`

`\downarrow`

`c)`

`( 34 - 2x ) * ( 2x - 6 ) = 0`

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

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

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

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

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

Vậy, `x \in {17; 3}`

`d)`

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

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

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

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

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

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

Vậy,` x \in {2019; 4}`

^{p/s: Bài này hnhu mk làm r mà ạ?}

1) Ta có: \(x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

Bài này giống tìm nghiệm quá :

1) \(5\left(x-7\right)=0\)

    \(\left(x-7\right)=0\div5\)

       \(\left(x-7\right)=0\)

     \(x=0+7\)

       \(x=7\)

2) \(25\left(x-4\right)=0\)

          \(\left(x-4\right)=0\div25\)

           \(\left(x-4\right)=0\)

              \(x=0+4\)

              \(x=4\)

3) \(34\left(2x-6\right)=0\)

    \(\left(2x-6\right)=0\div34\)

      \(\left(2x-6\right)=0\)

        \(2x=0+6\)

          \(2x=6\)

           \(x=6\div2\)

          \(x=3\)

4) \(2007\left(3x-12\right)=0\)

               \(\left(3x-12\right)=0\div2007\)

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

                   \(3x=0+12\)

                    \(3x=12\)

                     \(x=12\div3\)

                       \(x=4\)

5) \(47\left(5x-15\right)=0\)

         \(\left(5x-15\right)=0\div47\)

           \(\left(5x-15\right)=0\)

            \(5x=0+15\)

              \(5x=15\)

                 \(x=15\div5\)

                 \(x=3\)

6) \(13\left(4x-24\right)=0\)

         \(\left(4x-24\right)=0\div13\)

          \(\left(4x-24\right)=0\)

             \(4x=0+24\)

            \(4x=24\)

              \(x=24\div4\)

                \(x=6\)

a, 5(x – 7) = 0

x – 7 = 0

x = 7

b, 95 – 5(x+2) = 45

5(x+2) = 40

x+2 = 8

x = 6

c,  6 2 x + 2 3 + 40 = 100

6(2x+8) = 60

2x+8 = 10

x = 1

d, 3(3x+9)+6 = 96

3(3x+9) = 90

3x+9 = 30

3x = 27

x = 9

e,  2 6 + 5 + x = 3 4

5+x = 81–64

5+x = 17

x = 12

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