\(a,x+\dfrac{5}{7}=\dfrac{10}{-35}\\ x=\dfrac{-10}{35}-\dfrac{5}{7}\\ x=-1\\ b,\dfrac{2}{x}=\dfrac{x}{8}\\ x.x=2.8\\ x^2=16\\ x=\pm4\)
a, \(x\) + \(\dfrac{5}{7}\) = \(\dfrac{10}{-35}\)
\(x\) = \(-\dfrac{10}{35}\) - \(\dfrac{5}{7}\)
\(x\) = -1
b, \(\dfrac{2}{x}\) = \(\dfrac{x}{8}\)
\(x^2\) = 16
\(\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -4; 4}
c, \(\dfrac{x-3}{2}\) = \(\dfrac{32}{x-3}\)
(\(x-3\))(\(x-3\)) = 32\(\times\) 2
(\(x-3\))(\(x\) - 3) = 64
(\(x-3\))2 = 82
\(\left[{}\begin{matrix}x-3=-8\\x-3=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-8+3\\x=8+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=11\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -5; 11}