a) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Rightarrow72-20x-36x-84=30x-240-6x+84\)

\(\Rightarrow\left(72-84\right)-\left(20x+36x\right)=\left(30x-6x\right)-240+84\)

\(\Rightarrow-12-56=24x-56x\)

\(\Rightarrow-12+156=24x+56x\)

\(\Rightarrow144=80x\)

\(\Rightarrow x=144:80\)

\(\Rightarrow x=\frac{9}{5}\)

b) \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)

\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow15x+25-8x+12-5x-6x-36-1=0\)

\(\Rightarrow-4x=0\)

\(\Rightarrow-4.0\)

\(\Rightarrow x=0\)

à) bằng 1,8

b) bằng 0

a) 4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x +14)

<=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

<=> -20x - 36x - 30x + 6x = -72 - 84 - 240 - 84

<=> -80x = -480

<=> x = 6

b) 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1

<=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1

<=> 15x - 8x - 5x - 6x = -25 - 12 + 36 + 1

<=> -4x = 0

<=> x = 0

CTV sai dấu rồi kìa '-'

bài 1:

a) 1/3x=-5/3

x=-5

b) x+12/5=37/15

x=1/15

c) x-1/7=-36/7

x=-5

d) 3x-1/2=0

3x=1/2

x=1/6

e) 2/5+x=1/4

x=-3/20

Viết rõ ra hộ mình với

1/3x=4/3-3=-5/3

x=-5/3:1/3=-5

Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16

7+x=-16

    x=-16-7

    x=-23
2) 2x – 35 = 15

2x=15+35

2x=50

  x=50:2

  x=25
3) 3x + 17 = 12

3x=12-17

3x=-5

  x=-5/3
4) (2x – 5) + 17 = 6

2x-5=6-17

2x-5=-11

2x=-11+5

2x=-6

  x=-6:2

  x=-3
5) 10 – 2(4 – 3x) = -4

2(4-3x)=10-(-4)

2(4-3x)=14

4-3x=14:2

4-3x=7

3x=4-7

3x=-3

  x=-3:3

  x=-1
6) - 12 + 3(-x + 7) = -18

3(-x+7)=-18-(-12)

3(x+7)=-6

x+7=-6:3

x+7=-2

    x=-2-7

    x=-9

tự đi mà làm

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

TH1:x < 0 . PT có dạng

       \(-4\left(3x-1\right)-x+2\left(x-5\right)-7\left(x-3\right)=12\)

        \(4-12x-x+2x-10-7x+21=12\)

          \(15-18x=12\)

            \(x=\frac{1}{6}\left(koTM\right)\)

TH2:\(0\le x\le\frac{1}{3}\) PT có dạng:

       \(x-4\left(3x-1\right)-7\left(x-3\right)+2\left(x-5\right)=12\)

        \(x=\frac{3}{16}\left(TM\right)\)

TH3:\(\frac{1}{3}\le x< 3\) PT có dạng:

          \(x+4\left(3x-1\right)-7\left(x-3\right)+2\left(x-5\right)=12\)

           \(x=\frac{5}{8}\left(TM\right)\)

TH4:\(3\le x< 5\) PT có dạng:

            \(4\left(3x-1\right)+x+2\left(x-5\right)+7\left(x-3\right)=12\)

             \(x=\frac{47}{22}\left(koTM\right)\)

\(TH5:x\ge5\)PT có dạng:

             \(4\left(3x-1\right)+x-2\left(x-5\right)+7\left(x-3\right)=12\)

              \(x=1,5\left(koTM\right)\)

Vậy nghiệm PT là \(\frac{3}{16};\frac{5}{8}\)

a) 

<=> 3x - 3 + x - 2 = 2x - 2 - x + 1

<=> 3x + x - 2x + x = -2 + 1 + 3 + 2

<=>    3x               = 4

<=>   x                  = 4/3

Các câu sau làm tương tự

\(\left(3x-3\right)+\left(x-2\right)=\left(2x-2\right)-\left(x-1\right)\)

<=>   \(3x-3+x-2=2x-2-x+1\)

<=>  \(4x-5=x-1\)

<=> \(3x=4\)

<=>  \(x=\frac{4}{3}\)

Vậy....

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

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