mn ghi giúp em chi tiết bài giải nx ạ! 

đây nha

1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)

\(\Leftrightarrow5x+1-2x+4=4\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\dfrac{1}{3}\)

2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)

\(\Leftrightarrow9x+27+12-36x=-2x+2\)

\(\Leftrightarrow-27x+2x=2-39\)

hay \(x=\dfrac{37}{25}\)

3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x+6-10x=4-4x\)

\(\Leftrightarrow-7x+4x=4-6=-2\)

hay \(x=\dfrac{2}{3}\)

4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)

\(\Leftrightarrow5x-15-x-1=2x-4\)

\(\Leftrightarrow4x-2x=-4+16=12\)

hay x=6

5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)

\(\Leftrightarrow12x+3-9x+5+4x-8=0\)

\(\Leftrightarrow7x=0\)

hay x=0

\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\)  - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)

\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)

\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\) 

\(\dfrac{3}{16}\)  - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)

 ( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\) 

    \(\dfrac{41}{16}\)              - \(x\)    = \(\dfrac{5}{2}\)

                         \(x\)    = \(\dfrac{41}{16}\)  - \(\dfrac{5}{2}\)

                         \(x\)     = \(\dfrac{1}{16}\)

2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\)) 

    \(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)

   - \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)

    - \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)

       \(x\)   = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)

       \(x\)    = \(\dfrac{5}{6}\)

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`

`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`

`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`

`=> 3/16 - x + 19/8 = 2 1/2`

`=> 3/16 - x = 2 1/2 - 19/8`

`=> 3/16 - x =1/8`

`=> x = 3/16 - 1/8`

`=> x = 1/16`

Vậy, `x = 1/16`

`2,`

`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`

`=> 1/2 * (-11/15) = 1/5 - x + 4/15`

`=> -11/30 = x + 1/5 - 4/15`

`=> x + (-1/15) = -11/30`

`=> x = -11/30 + 1/15`

`=> x = -3/10`

Vậy, `x = -3/10.`

\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1989}{1991}\)

\(\Rightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{3980}{1991}.\dfrac{1}{2}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)

\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{1990}{1991}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1991}\)

\(\Rightarrow x+1=1991\)

\(\Rightarrow x=1990\)

Cho em cái đề bài ạ

Tìm x biết 

a: =>x-3=9

=>x=12

b: =>10-x=-26

=>x=36

c: =>x:4-1=2

=>x:4=3

=>x=12

d: =>x^2=4

=>x=2 hoặc x=-2

e: =>(x-2)^2=100

=>x-2=10 hoặc x-2=-10

=>x=12 hoặc x=-8

a)4/5+x=2/3

x=2/3-4/5

x=-2/15

b)-5/6-x=2/3

x=-5/6-2/3

x=-3/2

c)1/2x+3/4=-3/10

1/2x=-3/10-3/4

1/2x=-21/20

x=-21/20:1/2

x=-21/10

d)x/3-1/2=1/5

x/3=1/5+1/2

x/3=7/10

10x/30=21/30

10x=21

x=21:10

x=21/10