Question

Tìm x biết

a) [2/3-x]:3/4=1/5

b)1/3+2/3:x=-7

c) (-2x+3/5)²-9/25=0

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Answer 1

a; [\(\dfrac{2}{3}\) - \(x\)]: \(\dfrac{3}{4}\) = \(\dfrac{1}{5}\)

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

     \(\dfrac{2}{3}\) - \(x\)       = \(\dfrac{3}{20}\)

           \(x\)         =  \(\dfrac{2}{3}\) - \(\dfrac{3}{20}\)

           \(x\)         = \(\dfrac{31}{60}\)

Answer 2

b; \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\): \(x\) = -7

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

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

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

                   \(x\)  = - \(\dfrac{1}{11}\)

Answer 3

c; 

(-2\(x\) + \(\dfrac{3}{5}\))² - \(\dfrac{9}{25}\) = 0

 (\(\dfrac{3}{5}\) - 2\(x\))²          = \(\dfrac{9}{25}\)

\(\left[{}\begin{matrix}\dfrac{3}{5}-2x=\dfrac{-3}{5}\\\dfrac{3}{5}-2x=\dfrac{3}{5}\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=\dfrac{3}{5}+\dfrac{3}{5}\\2x=\dfrac{3}{5}-\dfrac{3}{5}\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=\dfrac{6}{5}\\2x=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{6}{5}:2\\x=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=0\end{matrix}\right.\)

Vậy \(x\) \(\in\) {0; \(\dfrac{3}{5}\)}