a; 5\(x\) -  7 = 3\(x\) + 9

    5\(x\)  - 3\(x\) = 9 + 7

    2\(x\)          = 16

      \(x\)           = 16: 2

       \(x\)          = 8

Vậy \(x=8\)

b; 1\(\dfrac{3}{4}\)\(x\) + 1\(\dfrac{1}{2}\) = - \(\dfrac{4}{5}\)

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

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

       \(\dfrac{7}{4}\)\(x\)         = - \(\dfrac{23}{10}\)

          \(x\)        = - \(\dfrac{23}{10}\) : \(\dfrac{7}{4}\)

          \(x\)       = - \(\dfrac{46}{35}\)

         Vậy \(x=-\dfrac{46}{35}\)

c; \(x\) + \(\dfrac{1}{2}\) = 2⁵:2³
    \(x\) + \(\dfrac{1}{2}\) = 2²

    \(x\) + \(\dfrac{1}{2}\) = 4

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

     \(x\)       = \(\dfrac{7}{2}\)

    Vậy \(x=\dfrac{7}{2}\)

d; (\(x+\dfrac{1}{2}\))² = \(\dfrac{4}{25}\) 

     ^{\(\left[{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{2}{5}\\x+\dfrac{1}{2}=\dfrac{2}{5}\end{matrix}\right.\)}

      ^{\(\left[{}\begin{matrix}x=-\dfrac{2}{5}-\dfrac{1}{2}\\x=-\dfrac{2}{5}+\dfrac{1}{2}\end{matrix}\right.\)}

      ^{\(\left[{}\begin{matrix}x=-\dfrac{9}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)}

vậy \(x\) \(\in\) {- \(\dfrac{9}{10}\); - \(\dfrac{1}{10}\)}

2.(\(\dfrac{1}{4}\) - 3\(x\)) = \(\dfrac{1}{5}\) - 4\(x\)

\(\dfrac{1}{2}\) - 6\(x\)      = \(\dfrac{1}{5}\) - 4\(x\)

- 4\(x\) + 6\(x\)   =\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) 

  2\(x\)         =  \(\dfrac{3}{10}\)

    \(x\)         =  \(\dfrac{3}{10}\): 2

   \(x=\dfrac{3}{20}\)

Vậy \(x=\dfrac{3}{20}\)

ai giải hộ tui vs tui tick cho

Cho mình hỏi (x-y)^2 (x+y)^2 hay ( x-y)^2 + (x+y)^2 ạ?