Toán Tìm x biết 2^x+2^x+1+2^x+2+……+2^x+2015=2^2019-8 04/07/2021 By Madelyn Tìm x biết 2^x+2^x+1+2^x+2+……+2^x+2015=2^2019-8
$2^x+2^{x+1}+2^{x+2}+…+2^{x+2015}=2^{2019}-8$ $⇒2^x(1+2+2^2+…+2^{2015})=2^{2019}-8$ $⇒2^x(2^{2016}-1)=2^{2019}-8$ $⇒2^x=\frac{2^{2019}-8}{2^{2016}-1}$ $⇒2^x=$$\frac{2^3.(2^{2016}-1)}{2^{2016}-1}$ $⇒2^x=2^3$ $⇒x=3$ Vậy $x=3$ Trả lời
Đáp án:x=3 Giải thích các bước giải: $2^{x}$+$2^{x+1}$+$2^{x+2}$+…+$2^{x+2015}$=$2^{2019}$-8 $2^{x}$(1+2+$2^{2}$+$2^{3}$+…+$2^{2015}$)=$2^{2019}$-$2^{3}$ $2^{x}$($2^{2016}$-1)=$2^{3}$($2^{2016}$-1) x=3 Trả lời
$2^x+2^{x+1}+2^{x+2}+…+2^{x+2015}=2^{2019}-8$
$⇒2^x(1+2+2^2+…+2^{2015})=2^{2019}-8$
$⇒2^x(2^{2016}-1)=2^{2019}-8$
$⇒2^x=\frac{2^{2019}-8}{2^{2016}-1}$
$⇒2^x=$$\frac{2^3.(2^{2016}-1)}{2^{2016}-1}$
$⇒2^x=2^3$
$⇒x=3$
Vậy $x=3$
Đáp án:x=3
Giải thích các bước giải:
$2^{x}$+$2^{x+1}$+$2^{x+2}$+…+$2^{x+2015}$=$2^{2019}$-8
$2^{x}$(1+2+$2^{2}$+$2^{3}$+…+$2^{2015}$)=$2^{2019}$-$2^{3}$
$2^{x}$($2^{2016}$-1)=$2^{3}$($2^{2016}$-1)
x=3