`2x-2/3=1/2`

`2x=1/2+2/3`

`2x=7/6`

`x=7/6:2=7/12`

\(2x-\dfrac{2}{3}=\dfrac{1}{2}\Leftrightarrow2x=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{7}{6}:2=\dfrac{7}{12}\)

2x - 2/3 = 1/2

2x = 1/2 + 2/3 = 7/6

x = 7/6 : 2 = 7/12

vậy x = 7/12

a,5x+8=2x-7                         

5x+8-2x+7=0

<=>3x+15=0

<=>3x=-15

<=>x=-5

Vậy x=-5

   b,3.(x+2)=2.(x-1)

<=>3x+6=2x-1

<=>3x+6-2x+1=0

<=>x+7=0

<=>x=-7

Vậy x=-7

a,5x+8=2x-7

   5x - 2x = -7 - 8

    3x        = -15

      x        = (-15) : 3

      x        = -5

Vậy x = -5

 b,3.(x+2)=2.(x-1)

    3x + 6 = 2x - 2

    3x - 2x= -2 - 6

       x     = -8

Vậy x = -8

# HOK TỐT #

\(a,5x+8=2x-7\)

<=>\(5x-2x=\left(-7\right)-8\)

<=>\(3x=-15\)

<=>\(x=-5\)

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

<=>\(3x+6=2x-2\)

<=>\(3x-2x=\left(-2\right)-6\)

<=>\(x=-8\)

 chọn lộn môn

Em đăng bài quả môn toán nhận hỗ trợ nhanh nhất nha

\(6\cdot x+9=15\)

\(\Rightarrow6\cdot x=15-9\)

\(\Rightarrow6\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{6}=1\)

_______________

\(43+2x=49\)

\(\Rightarrow2x=49-43\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=\dfrac{6}{2}=3\)

____________

\(x:2+5=11\)

\(\Rightarrow x:2=11-5\)

\(\Rightarrow x:2=6\)

\(\Rightarrow x=6\cdot2=12\)

______________

\(77-11x=0\)

\(\Rightarrow11x=77\)

\(\Rightarrow x=\dfrac{77}{11}\)

\(\Rightarrow x=7\)

_______________

\(12-4:x=8\)

\(\Rightarrow4:x=12-8\)

\(\Rightarrow4:x=4\)

\(\Rightarrow x=\dfrac{4}{4}=1\)

_____________

\(x:3+8=11\)

\(\Rightarrow x:3=11-8\)

\(\Rightarrow x:3=3\)

\(\Rightarrow x=3\cdot3=9\)

a: 6x+9=15

=>6x=6

=>x=1

b: 2x+43=49

=>2x=6

=>x=3

c: x:2+5=11

=>x:2=6

=>x=12

d: 77-11x=0

=>7-x=0

=>x=7

e: 12-4:x=8

=>4:x=4

=>x=1

f: x:3+8=11

=>x:3=3

=>x=9

a.x=1

b.x=3

c.x=12

d.x=7

e.x=1

f.x=9

\(a,\Rightarrow3x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ c,Đề.sai\\ d,Sửa:\left(x-2\right)^2-16\left(5-2x\right)^2=0\\ \Rightarrow\left[x-2-4\left(5-2x\right)\right]\left[x-2+4\left(5-2x\right)\right]=0\\ \Rightarrow\left(x-2-20+8x\right)\left(x-2+20-8x\right)=0\\ \Rightarrow\left(9x-22\right)\left(18-7x\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{9}\\x=\dfrac{18}{7}\end{matrix}\right.\)

\(\left(2x-1\right)^3-8\left(x-1\right)\left(x^2+x+1\right)+12x^2=2x+1\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8\left(x^3-1\right)+12x^2-2x-1=0\)

\(\Leftrightarrow4x+6=0\)

\(\Leftrightarrow2\left(2x+3\right)=0\)

\(\Leftrightarrow2x=-3\)

\(\Leftrightarrow x=\frac{-3}{2}\)

đề sai nhé

phải là

8. (x-1) . (x²+x+1)

I 2x-3 I = I x+1 I

2x-3 = x+1

x+1 - 2x+3=0

x (1-2) +1+3=0

-1x +4 =0

-1x      = 0-4

-1x      =-4

x          = -4 : -1

x         =4

Trả lời:

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

\(\Rightarrow2x-3=x+1\) hoặc   \(2x-3=-\left(x+1\right)\)

TH1:   \(2x-3=x+1\)

           \(2x-x=1+3\)

            \(x=4\)

TH2: \(2x-3=-\left(x+1\right)\)

         \(2x-3=-x-1\)

          \(2x+x=-1+3\)

          \(3x=2\)

          \(x=\frac{2}{3}\)

          Vậy \(x=4;x=\frac{2}{3}\)

\(\left|\frac{1}{3}-x\right|-\left|-\frac{2}{3}\right|=\left|\frac{3}{4}\right|\)

\(\Leftrightarrow\frac{1}{3}-x-\frac{2}{3}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{3}-x=\frac{3}{4}+\frac{2}{3}\)

\(\Leftrightarrow\frac{1}{3}-x=\frac{17}{12}\)

\(\Leftrightarrow x=\frac{1}{3}-\frac{17}{12}=\frac{4}{12}-\frac{17}{12}=\frac{-13}{12}\)

a) \(\left|2x-1\right|+\frac{1}{3}=0\)

\(\Leftrightarrow\left|2x-1\right|=-\frac{1}{3}\)

=> vô lý

=> PT vô nghiệm

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

\(\Leftrightarrow\left|x+2\right|=-\left|x-3\right|\)

Vì \(\hept{\begin{cases}\left|x+2\right|\ge0\\-\left|x-3\right|\le0\end{cases}\left(\forall x\right)}\) nên dấu "=" xảy ra khi: 

\(\left|x+2\right|=-\left|x-3\right|=0\Rightarrow\hept{\begin{cases}x=-2\\x=3\end{cases}}\) (vô lý)

=> PT vô nghiệm