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Lời giải:
$A-4=1+4+4^2+4^3+...+4^{2008}$
$4(A-4)=4+4^2+4^3+...+4^{2009}$
$\Rightarrow 4(A-4)-(A-4)=4^{2009}-1$
$\Rightarrow 3(A-4)=4^{2009}-1$
$\Rightarrow 3A=4^{2009}+11> 4^{2009}=4.4^{2008}$
$\Rightarrow A> \frac{4.4^{2008}}{3}> 4^{2008}$
$\Rightarrow 2A> 2.4^{2008}> 4^{2006}$ hay $2A> B$
Hay $A> \frac{B}{2}$
Đề sai bạn xem lại.
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\(2^x+2^{x-1}+2^{x-2}=2^{x-2}.2^2+2^{x-2}.2+2^{x-2}\)\(=2^{x-2}\left(4+2+1\right)=2^{x-2}.7\)
thank you
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1) 41 - 2x + 1 = 9
-2x + 1 = 9 - 41
-2x + 1 = -32
2x + 1 = 32
2x + 1 = 25
x + 1 = 5
x = 5 - 1
x = 4
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\(2^{x+1}-2^x=32\)
\(\Rightarrow2^x.2-2^x=32\)
\(\Rightarrow2^x\left(2-1\right)=32\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x=5
Hok tốt !!!!!!!!!!!!!!!!!
2x+1 - 2x = 32
⇔ 2x( 2 - 1 ) = 32
⇔ 2x . 1 = 32
⇔ 2x = 32
⇔ 2x = 25
⇔ x = 5
Vậy x = 5
(x-1)2 = 1
(x-1)2 = 12
=> x-1=1
x=1+1
x=2
Vậy x=2
_HT_