Question

(2²⁰¹⁹ +2²⁰²⁰ + 2²⁰²¹) / x = 32⁴⁰³/0.5;
x=?​

Answer 1

Explicație pas cu pas:

$\frac{ {2}^{2019} + {2}^{2020} + {2}^{2021} }{x} = \frac{ {32}^{403} }{0.5}$

$\frac{ {2}^{2019}(1 + 2 + {2}^{2}) }{x} = \frac{ {( {2}^{5} )}^{403} }{ \frac{1}{2} }$

$\frac{ {2}^{2019}(1 + 2 + 4) }{x} = \frac{ {{2}^{2015}} }{ {2}^{ - 1} }$

$\frac{ 7 \cdot {2}^{2019} }{x} = {2}^{2016} \iff x = \frac{7 \cdot {2}^{2019}}{ {2}^{2016} }$

$x = 7 \cdot {2}^{2019 - 2016} = 7 \cdot {2}^{3} = 7 \cdot 8 \\ \implies \red{\bf x = 56}$

Answer 2

$\it 2^{2019}+2^{2020}+2^{2021}=2^{2016}(2^3+2^4+2^5)=2^{2016}\cdot(8+16+32)=\\ \\ =2^{2016}\cdot56\\ \\ \\ \dfrac{32^{403}}{0,5}=\dfrac{(2^5)^{403}}{\dfrac{1}{2}}=2^{2015}\cdot2=2^{2016}$

Ecuația devine:

$\it \dfrac{2^{2016}\cdot56}{x}=2^{2016}\Big|_{:2^{2016}} \Rightarrow \dfrac{56}{x}=1 \Rightarrow x=56$