a; \(\dfrac{x}{6}\) = \(\dfrac{-3}{4}\)

     \(x=\dfrac{-3}{4}.6\)

     \(x\) = - \(\dfrac{9}{2}\)

Vậy \(x=-\dfrac{9}{2}\)

b; \(\dfrac{5}{x}\) = \(\dfrac{15}{-20}\) (đk \(x\ne0\))

    \(x\) = 5 : \(\dfrac{15}{-20}\)

     \(x=-\dfrac{20}{3}\)

Vậy \(x=-\dfrac{20}{3}\)

c; \(\dfrac{x+11}{14-x}\) = \(\dfrac{2}{3}\) (đk \(x\ne14\))

     3.(\(x+11\)) = 2.(14 - \(x\))

    3\(x\) + 33 = 28 - 2\(x\)

     3\(x\) + 2\(x\) =  28 - 33

         5\(x\)    = -5

            \(x\)    = -1

Vậy \(x\)        = -1 

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?

Ta có:

\(y=2x\)

\(x=\dfrac{1}{3}z\)

\(\Rightarrow y=2\left(\dfrac{1}{3}z\right)\)

\(\Rightarrow y=\dfrac{2}{3}z\) 

Vậy y tỉ lệ thuận với z theo hệ số tỉ lệ \(\dfrac{2}{3}\).