Question

[Thủy Tiên]

 

Answer 1

[Nguyễn Thị Thương Hoài]

a, \(\sqrt{\left(\dfrac{x}{5}-1\right)^2}\) = \(\dfrac{4}{3}\)

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

             \(\left[{}\begin{matrix}\dfrac{x}{5}-1=\dfrac{-4}{3}\left(x< 5\right)\\\dfrac{x}{5}-1=\dfrac{4}{3}\left(x>5\right)\end{matrix}\right.\)  

              \(\left[{}\begin{matrix}\dfrac{x}{5}=-\dfrac{1}{3}\\\dfrac{x}{5}=\dfrac{7}{3}\end{matrix}\right.\)

                \(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{35}{3}\end{matrix}\right.\)

vậy \(x\) \(\in\) { - \(\dfrac{5}{3}\); \(\dfrac{35}{3}\)}

Answer 2

[Nguyễn Thị Thương Hoài]

b, \(\sqrt{\left(7-x\right)\left(8+x\right)}\) = 0

        \(\left[{}\begin{matrix}7-x=0\\8+x=0\end{matrix}\right.\)

        \(\left[{}\begin{matrix}x=7\\x=-8\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-8; 7}