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A= \(2^0+2^1+2^2+...+2^{50}\)
\(\Rightarrow\)2A =2(\(2^0+2^1+2^2+...+2^{50}\))
\(\Rightarrow\)2A= \(2+2^2+2^3+2^4+...+2^{51}\)
\(\Rightarrow\)2A-A= (\(2+2^2+2^3+2^4+...+2^{51}\))-(\(2+2^2+2^3+2^4+...+2^{50}\))
\(\Rightarrow\)A= \(2^{51}-1\)
\(2^{x+1}+5.2^{x+2}=88\)
\(\Rightarrow2^x.2+5.2^x.2^2=88\)
\(\Rightarrow2^x\left(2+5.2^2\right)=88\)
\(\Rightarrow2^x.22=88\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
Vậy x = 2
Theo bài ra ta có:
2x+1+5.2x+2 = 88
=> 2x.2+5.2x.22
=> 2x.( 2 = 5.22 )
=> 2x.22 = 88
=> 2x = 88:22
=> 2x = 4
=> x = 4:2
=> x = 2
Vậy x = 2
Bài 1: \(5^3+3^5\)
=125+243
=368
Bài 2:
a: \(\left(x-1\right)^3=125\)
=>\(\left(x-1\right)^3=5^3\)
=>x-1=5
=>x=6
b: \(\frac{720}{41-\left(2x-5\right)}=2^3\cdot5\)
=>\(\frac{720}{41-\left(2x-5\right)}=8\cdot5=40\)
=>41-(2x-5)=720/40=18
=>2x-5=23
=>2x=28
=>x=14
Bài 3:
a: \(\frac19\cdot3^4\cdot3^{n}=3^7\)
=>\(3^{n+4-2}=3^7\)
=>n+4-2=7
=>n+2=7
=>n=5
b: \(\left(2^2:4\right)\cdot2^{n}=4\)
=>\(\frac44\cdot2^{n}=4\)
=>\(2^{n}=4=2^2\)
=>n=2
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{15}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{14}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{14}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{15}}\right)\)
\(A=1-\frac{1}{2^{15}}\)\
\(A=1-\frac{1}{32768}\)
\(A=\frac{32767}{32768}\)