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Ta có: \(A=4^0+4^1+4^2+...+4^{20}\)
Nhân A với 4 ta có:
\(4A=4\left(4^0+4^1+4^2+...+4^{20}\right)\)
=> \(4A-A=\left(4^1+4^2+4^3+...+4^{21}\right)-\left(4^0+4^1+4^2+...+4^{20}\right)\)
=> \(A\left(4-1\right)=4^{21}-4^0\)
=> \(3A=4^{21}-1\)
=> \(3A+1=4^{21}=\left(4^3\right)^7=64^7>63^7\)
Vậy 3A + 1 > 63^7.
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a: =>(2x+1/2)^2=1/4
=>2x+1/2=1/2 hoặc 2x+1/2=-1/2
=>x=-1/2 hoặc x=0
b: =>(x-1/5)^2=49
=>x-1/5=7 hoặc x-1/5=-7
=>x=-6,8 hoặc x=7,2
c: =>1,2x=12
=>x=10
d: =>3/4x+1/2x+1/2=-11/4
=>5/4x=-11/4-2/4=-13/4
=>x=-13/5
e: =>-0,25x+1,25x=0,2
=>x=0,2
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\(a.3x-4^{20}:4^{17}=4^0\\ 3x-4^3=1\\ 3x-64=1\\ 3x=64+1\\ 3x=65\\ x=\dfrac{65}{3}\)
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a, \(7^{2010}:7=7^{2009}\)
b, \(4^{20}:4^0=4^{20}\)
c, \(\frac{3^8.3^{15}}{3^7.3^{14}}=\frac{3^{23}}{3^{21}}=3^2\)
d, \(\frac{4^{16}.4^2}{4^{20}:4^2}=\frac{4^{18}}{4^{18}}=4^0\)
e, \(\frac{2^{15}:2^{10}}{2^{18}:2^{15}}=\frac{2^5}{2^3}=2^2\)
f, \(25^4:125=\left(5^2\right)^4:5^3=5^8:5^3=5^5\)
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a, 3 ( x + 1 ) - 2 ( 3 x - 4 ) = - 13
=> 3x + 3 - 6x + 8 = - 13
=> 6x - 3x = 3 + 8 + 13
=> 3x = 24
=> x = 8
b, 2 ( x - 3 ) - 4 ( 2 x - 1 ) = - 20
=> 2x - 6 - 8x + 4 = - 20
=> 8x - 2x = - 6 + 4 + 20
=> 6x = 18
=> x = 3
c, 2 x ( x + 3 ) = 0
=> \(\orbr{\begin{cases}2x=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)
d, ( x - 1 ) ( 5 x - x ) = 0
=> \(\orbr{\begin{cases}x-1=0\\5x-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\4x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)
e, ( x + 3 ) 2 ( 4 - x ) = 0
=> \(\orbr{\begin{cases}\left(x+3\right)^2=0\\4-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x+3=0\\4-x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
a) \(3\left(x+1\right)-2\left(3x-4\right)=-13\)
\(\Leftrightarrow3x+3-6x+8=-13\)
\(\Leftrightarrow3x-6x=-13-3-8\)
\(\Leftrightarrow-3x=-24\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
b) \(2\left(x-3\right)-4\left(2x-1\right)=-20\)
\(\Leftrightarrow2x-6-8x+4=-20\)
\(\Leftrightarrow2x-8x=-20+6-4\)
\(\Leftrightarrow-6x=-18\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
c) \(2x\left(x+3\right)=0\)
\(\orbr{\begin{cases}2x=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
d)\(\left(x-1\right)\left(5x-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
e)\(\left(x+3\right)^2\left(4-x\right)=0\)
\(\orbr{\begin{cases}\left(x+3\right)^2=0\\4-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+3=0\\-x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
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a*2 =2+4^2+4^3+...+4^20+4^21
a*2-a=4+4^21
4^21=4*4*4*...*4
=16^5*4+4 =...8 chia 5 du 3
\(A=4^0+4^2+4^3+...+4^{20}\)
\(\Rightarrow4^2A=4^2+4^4+4^6+..+4^{22}\)
\(\Rightarrow16A-A=4^2+4^4+...+4^{22}-4^0-4^2-4^4-...-4^{20}\)
\(\Rightarrow15A=4^{22}-4^0\)
\(\Rightarrow15A=4^{22}-1\)
\(\Rightarrow A=\dfrac{4^{22}-1}{15}\)
4^2A=4^2+4^4+4^6+....+4^22
16A - A= 4^22-4^0
15A=4^22-1
A=(4^22-1):15