Tìm x , biết : a, -32/(-2)^3=4 b,8/2^x-1=2 c,(1/2)^2x-1=1/8 d,x/4 / 2 = 4/ x/2 Mn giúp e zới ạ

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1. a) $\dfrac{-32}{(-2)^3}$

   $=\dfrac{-32}{-8}=4=VP$ (ĐPCM)

   b) $(\dfrac{8}{2})^{x-1}=2$

   $↔4^{x-1}=2$

   $↔(2^2)^{x-1}=2$

   $→2^{2x-2}=2$

   $↔2x-2=1$

   $↔2x=3$

   $↔x=\dfrac{3}{2}$

   c) $(\dfrac{1}{2})^{2x-1}=\dfrac{1}{8}$

   $↔(\dfrac{1}{2})^{2x-1}=(\dfrac{1}{2})^3$

   $↔2x-1=3$

   $↔2x=4$

   $↔x=2$

   d) $\dfrac{x}{4}:2=4:\dfrac{x}{2}$

   $↔\dfrac{x}{4}.\dfrac{1}{2}=4.\dfrac{2}{x}$

   $↔\dfrac{x}{8}=\dfrac{8}{x}$

   $↔x^2=8.8=64$

   \(↔\left[ \begin{array}{l}x=\sqrt{64}=8\\x=-\sqrt{64}=-8\end{array} \right.\) 

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2. Đáp án:

    

   Giải thích các bước giải:

   $a) ???$

   $b) \dfrac{8}{2^{x-1}}=2$

   $⇔2^{x-1}=8:2$

   $⇔2^{x-1}=4$

   $⇔2^{x-1}=2^2$

   $⇔x-1=2 ⇔ x=3$

   $c) \left (\dfrac{1}{2} \right )^{2x-1}=\dfrac{1}{8}$

   $⇔\left (\dfrac{1}{2} \right )^{2x-1}=\left (\dfrac{1}{2} \right )^3$

   $⇔2x-1=3$

   $⇔2x=4$

   $⇔x=2$

   $d) \dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}$

   $⇔\dfrac{x}{8}=\dfrac{8}{x}$

   $⇔x.x=8.8$

   $⇔x^2=64$

   $⇔x=±8$

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