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3) \(...\Rightarrow2^x\left(2^3+1\right)=36\)
\(\Rightarrow2^x.9=36\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\Rightarrow x=2\)
4) \(...\Rightarrow4^{x+1}-4^x=12\)
\(\Rightarrow4^x\left(4-1\right)=12\)
\(\Rightarrow4^x.3=12\)
\(\Rightarrow4^x=4=4^1\Rightarrow x=1\)
5) \(...\Rightarrow5^{x+1}\left(5^2-1\right)=3000\)
\(\Rightarrow5^{x+1}.24=3000\)
\(\Rightarrow5^{x+1}=125\)
\(\Rightarrow5^{x+1}=5^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
6) Bạn xem lại đề
a. \(2^x.2^3+2^x=36\)
\(2^x\left(2^3+1\right)=36\)
\(2^x.9=36\)
\(2^x=4\Rightarrow x=2\)
b. \(4^x.4^1-\left(2^2\right)^x=12\)
\(4^x.4-4^x=12\)
\(4^x\left(4-1\right)=12\)
\(4^x.3=12\)
\(4^x=4\)
x = 1
c. \(5^x.5^3-5^x.5^1=3000\)
\(5^x\left(5^3-5^1\right)=3000\)
\(5^x.120=3000\)
\(5^x=25\)
x = 2
d. \(4^{x+1}=2^{2x}\)
\(4^x.4=\left(2^2\right)^x\)
\(4^x.4=4^x\)
Có vẻ như câu 4 này để bài thiếu
Đặt \(A=2^{2x}+2^{2x+1}+...+2^{2x+1918}\)
=>\(2\cdot A=2^{2x+1}+2^{2x+2}+...+2^{2x+1919}\)
=>\(A=2^{2x+1919}-2^{2x}\)
Theo đề, ta có; \(2^{2x+1919}-2^{2x}=2^{1923}-2^4\)
=>\(2^{2x}\cdot\left(2^{2019}-1\right)=2^4\left(2^{2019}-1\right)\)
=>2x=4
=>x=2
22x + 20x + 18x + ... + 2x = 264
=> x(22 + 20 + 18 + ... + 2) = 264
=> x.132 = 264
=> x = 2
vậy_
a) \(\left(-5\right)^3.\left(-19\right).32.\left(-2018\right)^0\)
\(=\left(-125\right).\left(-608\right).1\)
\(=76000\)
b) \(\left(-195\right).56-43.195-195\)
\(=195.\left(-56-43-1\right)\)
\(=195.\left(-100\right)\)
\(=-19500\)
c) \(\left(-37\right).520+\left(-260\right).16-90-\left(-160\right)\)
\(=-19240+4160-90+160\)
\(=-15010\)
a, ( -5 )3 . ( -19 ) . 32 . ( -2018 )0
= -125 . ( -19 ) . 32 . 1
= 2375 . 32 . 1
= 76000
b, -195 . 56 - 43 . 195 - 195
= -10920 - 8385 - 195
= 2523 - 195
= 2340
\(2\cdot2^{2x}+4^3\cdot2^{x-1}=520\)
\(2^1\cdot2^{2x}+\left(2^2\right)^3\cdot2^{x+1}=520\)
\(2^{2x+1}+2^{2x+1}\cdot2^6=520\)
\(2^{2x+1}\cdot\left(1+2^6\right)=520\)
\(2^{2x+1}\cdot65=520\)
\(2^{2x+1}=520:65\)
\(2^{2x+1}=8\)
\(2^{2x+1}=2^3\)
\(2x+1=3\)
\(2x=3-1\)
\(2x=2\)
\(x=2:2\)
\(x=1\)