Đáp án:
$d) \dfrac{x+7}{-20} =\dfrac{-5}{x+7}$ (đk x $\neq$ 7)
$⇔(x+7)(x+7) = -5 .(-20)$
$⇔x^2+7x+7x+49 = 100$
$⇔x^2+14x+49-100=0$
$⇔x^2+14x-51=0$
$⇔x^2-3x+17x-51=0$
$⇔x(x-3)+17(x-3)=0$
$⇔(x-3)(x+17)=0$
$⇔$\(\left[ \begin{array}{l}x-3=0\\x+17=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=3(nhận)\\x=-17(nhận)\end{array} \right.\)
$\text{x ∈ { 3 ; -17}}$
$e) \dfrac{x}{-8}=\dfrac{2}{x^3}$ (đk x $\neq$ 0)
$⇔ -x^4 = -16$
$⇔x^4 =16$
$⇔x^2 =(±2)^4$
$⇔x=±2$ (nhận)
$\text{Vậy $x=±2$}$
$f) \dfrac{x}{8} =\dfrac{x}{x^3}$ (đk : x $\neq$ 0)
$⇔x^4 = 8x$
$⇔x^4-8x=0$
$⇔x(x^3-8)=0$
$⇔$\(\left[ \begin{array}{l}x=0\\x^3-8=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=0(loại)\\x=2(nhận)\end{array} \right.\)
$\text{Vậy x = 2}$
