2/3.x + 1/4 = 7/12

2/3.x           = 7/12 - 1/4

2/3.x           = 1/3

x                 = 1/3 : 2/3

x                 = 1/2

Bài làm

\(\frac{2}{3}x+\frac{1}{4}=\frac{7}{12}\)

\(\frac{2}{3}x=\frac{7}{12}-\frac{1}{4}\)

\(\frac{2}{3}x=\frac{7}{12}-\frac{3}{12}\)

\(\frac{2}{3}x=\frac{4}{12}\)

\(\frac{2}{3}x=\frac{1}{3}\)

\(x=\frac{1}{3}:\frac{2}{3}\)

\(x=\frac{1}{3}.\frac{3}{2}\)

\(x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

2/3 .x + 1/4 = 7/12

2/3 .x         = 7/12 - 1/4

2/3.x          = 1/3

x                = 1/3 : 2/3

x                = 1/2

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

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

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

\(C=\dfrac{-5}{7}+\dfrac{-2}{7}+\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{-1}{5}=-1+1-\dfrac{1}{5}=\dfrac{-1}{5}\)

\(\dfrac{7}{5}+\dfrac{4}{7}-\dfrac{9}{10}=\dfrac{49-20}{35}-\dfrac{9}{10}=\dfrac{19}{35}-\dfrac{9}{10}=\dfrac{190-315}{350}=\dfrac{-125}{350}\)

\(\dfrac{2}{1}+\dfrac{3}{4}\text{×}\dfrac{8}{5}=\dfrac{8+3}{4}\text{×}\dfrac{8}{5}=\dfrac{11\text{×}8}{4\text{×}5}=\dfrac{88}{20}\)

mấy câu kia áp dụng là dc!

c)(x-4).(2x+6)=0

=>(x-4)=0 hoặc (2x+6)=0

với x-4 = 0

      x    =0+4

      x    =4

với 2x+6=0

      2x    =0-6

      2x    =-6

      x      =-6:2

      x      =-3

3x=(-7)+1

3x=(-6)

x=(-6):3

x=(-2)

=(-5)=(-5).(-2) là sao vậy bạn

Bài b) (x-4)(x-7)(x-6)(x-5)=1680

=> (x²-11x+28)(x²-11x+30)=1680

Đặt t=x²-11x+28

=> t(t+2)=1680

=>t²+2t-1680=0

=> t²+2t+1-1681=0

=> (t+1)²-41²=0

=> (t-40)(t+42)=0

=> t=40 hoặc t=-42

Bạn thế vào như câu a) để giải nhé !!!

a.X=-3

b.X=-1

a.X=-3

b.x=-1

|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)

|7 - \(\dfrac{3}{4}x\)|  - \(\dfrac{3}{2}\) = 2

|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)

|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)

\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\) 

\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)

 5  - |\(x-3\)| = 5

       |\(x-3\)| = 5 - 5

      |\(x-3\)| = 0

      \(x-3\) = 0 

      \(x\) = 3

|\(\dfrac{7}{2}\) - 42| + 14 = 14 ( vô lý xem lại đề bài nhé em)

\(\dfrac{2}{3}\) - |\(x+2\)| = 2

     |\(x+2\)| = \(\dfrac{2}{3}\) - 2

    |\(x+2\)| = - \(\dfrac{4}{3}\)  

   |\(x+2\)| ≥ 0 ∀ \(x\)

Vậy \(x\in\) \(\varnothing\)