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
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
Share
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Tryphena
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) $\dfrac{x}{4}:2=4:\dfrac{x}{2}$
$↔\dfrac{x}{4}.\dfrac{1}{2}=4.\dfrac{2}{x}$
$↔\dfrac{x}{8}=\dfrac{8}{x}$
$↔x^2=8.8=64$
\(↔\left[ \begin{array}{l}x=\sqrt{64}=8\\x=-\sqrt{64}=-8\end{array} \right.\)
Amity
Đáp án:
Giải thích các bước giải:
$a) ???$
$b) \dfrac{8}{2^{x-1}}=2$
$⇔2^{x-1}=8:2$
$⇔2^{x-1}=4$
$⇔2^{x-1}=2^2$
$⇔x-1=2 ⇔ x=3$
$c) \left (\dfrac{1}{2} \right )^{2x-1}=\dfrac{1}{8}$
$⇔\left (\dfrac{1}{2} \right )^{2x-1}=\left (\dfrac{1}{2} \right )^3$
$⇔2x-1=3$
$⇔2x=4$
$⇔x=2$
$d) \dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}$
$⇔\dfrac{x}{8}=\dfrac{8}{x}$
$⇔x.x=8.8$
$⇔x^2=64$
$⇔x=±8$