a, $\dfrac{1}{2}+x=\dfrac{-1}{4}$
$⇒x=\dfrac{-1}{4}-\dfrac{1}{2}=\dfrac{-3}{4}$
Vậy: $x=\dfrac{-3}{4}$
b, `\frac{-9}{8}+\frac{-3}{8}.(x-\frac{2}{3})=\frac{-17}{16}`
`⇒\frac{-3}{8}.(x-\frac{2}{3})=\frac{1}{16}`
$⇒x-\dfrac{2}{3}=\dfrac{-1}{6}$
$⇒x=\dfrac{-1}{6}+\dfrac{2}{3}=\dfrac{1}{2}$
Vậy: $x=\dfrac{1}{2}$
c, $\dfrac{x-2}{-7}=\dfrac{-7}{x-2}$
$⇒\dfrac{-x+2}{7}=\dfrac{-7}{x-2}$
$⇒(-x+2)(x-2)=-49$
$⇒-x^2+4x+45=0$
$⇒(-x^2+9x)-(5x+45)=0$
$⇒-x(x-9)-5(x-9)=0$
$⇒(x-9)(-x-5)=0$
$⇒$\(\left[ \begin{array}{l}x-9=0\\-x-5=0\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x=9\\x=-5\end{array} \right.\)
Vậy: `x∈{9;-5}`
d, `|\frac{5}{8}-x|=\frac{10}{-6}:\frac{5}{-3}`
`⇒|\frac{5}{8}-x|=1`
$⇒$\(\left[ \begin{array}{l}\dfrac{5}{8}-x=1\\\dfrac{5}{8}-x=-1\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x=\dfrac{13}{8}\\x=\dfrac{-3}{8}\end{array} \right.\)
Vậy: `x∈{\frac{13}{8};\frac{-3}{8}}`
