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bài này dễ lắm

\(2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-8\)

\(\Leftrightarrow2^x\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-2^3\)

\(\Leftrightarrow2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)\)

\(\Leftrightarrow2^x=2^3\)

\(\Leftrightarrow x=3\)

Vậy x = 3

2 ^x + 2^x+1+ 2 ^x+2+.......+ 2^x+2015=2²⁰¹⁹-8

=2^x.( 1+2+2²+2³+.....+ 2 ²⁰¹⁵)=2²⁰¹⁹- 2³

đặt A= 1+2+2²+...+2²⁰¹⁵

=>2A=2+2²+2³+..+2²⁰¹⁶

=>2A -A = ( 2+ 2²+2³+......+2²⁰¹⁶)-(1+2+2²+........+2²⁰¹⁵)=A=2²⁰¹⁶-1

\(\Rightarrow\)2^x.(2²⁰¹⁶-1)=2³.(2²⁰¹⁶-1)

=>x=3

a) \(\Rightarrow2^x.1+2^x.2+2^x.3+...+2^x.2015=2^{2019}-2^3\)

\(\Rightarrow2^x.\left(1+2+2^2+2^3+...+2^{2015}\right)=2^3.\left(2^{2016}-1\right)\)

Đặt \(A=1+2+2^2+2^3+...+2^{2015}\)

\(\Rightarrow2.A=2+2^2+2^3+...+2^{2016}\)

\(\Rightarrow A=2^{2016}-1\)

\(\Rightarrow2^x.\left(2^{2016}-1\right)=2^3.\left(2^{2016}-1\right)\)

\(\Rightarrow2^x=2^3\Rightarrow x=3\)

Vậy x=3

Lời giải:

$A=2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}$
$2A=2^{x+1}+2^{x+2}+....+2^{x+2016}$

$2A-A=2^{x+2016}-2^x$

$A=2^{x+2016}-2^x=2^x(2^{2016}-1)$

Vậy:

$2^x(2^{2016}-1)=2^{2019}-8=2^{2019}-2^3=2^3(2^{2016}-1)$

$\Rightarrow 2^x=2^3$
$\Rightarrow x=3$

Theo đầu bài ta có:
\(2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-8\)
\(\Rightarrow2\left(2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}\right)-\left(2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}\right)=2^{2019}-8\)
\(\Rightarrow\left(2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2016}\right)-\left(2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}\right)=2^{2019}-8\)
\(\Rightarrow2^{x+2016}-2^x=2^{2019}-8\)
\(\Rightarrow2^x\cdot2^{2016}-2^x=2^3\cdot2^{2016}-2^3\)
\(\Rightarrow2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)\)
\(\Rightarrow2^x=2^3\)
\(\Rightarrow x=3\)

cảm ơn

Ta có :

a) \(1+3+5+...+\left(2x-1\right)=\frac{\left(2x-1\right)+1}{2}\left(\frac{\left(2x-1\right)-1}{2}+1\right)=x^2\)

\(\Leftrightarrow x^2=225\Rightarrow x=15\)

b) \(2^x+2^{x+1}+...+2^{x+2015}=2^x\left(2^0+2^1+...+2^{2015}\right)\)

Đặt A = 2⁰ + 2¹ + ... + 2²⁰¹⁵ . Ta có :

2A = 2¹ + 2² + ... + 2²⁰¹⁶

⇒ A = 2A - A = (2¹ +2² +...+2²⁰¹⁶ )-(2⁰ + 2¹ + ... +2²⁰¹⁵ )

⇒ A = 2²⁰¹⁶ - 1

⇔ 2^x.A = 2²⁰¹⁹ - 8

⇔ 2^x( 2²⁰¹⁶ - 1 ) = 2³ ( 2²⁰¹⁶ - 1 )

⇔ x = 3

Đề bài c) chưa đủ ý nên o làm đc

a) 

b) 

Đặt A = 2⁰ + 2¹ + ... + 2²⁰¹⁵ . Ta có :

2A = 2¹ + 2² + ... + 2²⁰¹⁶

⇒ A = 2A - A = (2¹ +2² +...+2²⁰¹⁶ )-(2⁰ + 2¹ + ... +2²⁰¹⁵ )

⇒ A = 2²⁰¹⁶ - 1

⇔ 2^x.A = 2²⁰¹⁹ - 8

⇔ 2^x( 2²⁰¹⁶ - 1 ) = 2³ ( 2²⁰¹⁶ - 1 )

⇔ x = 3

minh chua lam dc cau c

x=3 nha

k cho mk nha

x=3 nha

k mk nha