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\(4^x+10.4^x=1280\)
<=>\(11.4^x=1280\)
<=>\(4^x=\frac{1280}{11}=>saiđề\)
1280:[109-(3x-7)]=40
109-(3x-7)=1280/40
109-(3x-7)=32
3x-7=109-32
3x-7=77
3x=77+7
3x=84
x=84/3
\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)
\(=\frac{1}{5}\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(=\frac{\frac{1}{5}\left(1-\frac{1}{2^9}\right)}{\left(1-\frac{1}{2}\right)}\)
\(=\frac{2}{5}\left(1-\frac{1}{2^9}\right)\)
\(A=\dfrac{1}{2^0.5}+\dfrac{1}{2^1.5}+\dfrac{1}{2^2.5}+...+\dfrac{1}{2^8.5}\)
\(5A=\dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^8}\)
\(5A=2-1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+...++\dfrac{1}{128}+\dfrac{1}{256}\)
\(5A=2-\dfrac{1}{256}=\dfrac{511}{256}\)
\(A=\dfrac{511}{1280}\)
C = \(\frac{1}{5}\)+\(\frac{1}{10}\)+\(\frac{1}{20}\)+\(\frac{1}{40}\)+\(\frac{1}{80}\)+........+\(\frac{1}{1280}\)
2C = 2 . ( \(\frac{1}{5}\)+\(\frac{1}{10}\)+.......+\(\frac{1}{1280}\))
2C = \(\frac{2}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{10}\)+.....+\(\frac{1}{1280}\)
2C-C = ( \(\frac{2}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{10}\)+......+\(\frac{1}{1280}\)) - (\(\frac{1}{5}\)+\(\frac{1}{10}\)+.....+\(\frac{1}{1280}\))
C . ( 2-1) = \(\frac{2}{5}\)
C = \(\frac{2}{5}\)
Vậy C = \(\frac{2}{5}\)
\(C=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+\frac{1}{80}+........+\frac{1}{1280}\)
\(\Rightarrow2C=2\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+\frac{1}{80}+...........+\frac{1}{1280}\right)\)
\(\Rightarrow2C=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+.............+\frac{1}{1280}\)
\(\Rightarrow2C-C=\left(\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+............+\frac{1}{1280}\right)-\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+\frac{1}{80}+...........+\frac{1}{1280}\right)\)
\(\Rightarrow C=\frac{2}{5}-\frac{1}{1280}\)
\(\Rightarrow C=\frac{512}{1280}-\frac{1}{1280}\)
\(\Rightarrow C=\frac{511}{1280}\)
Vậy C = \(\frac{511}{1280}\)
a: \(\left\{100:\left[45-45:\left(-9\right)\right]-32\right\}\cdot5-2\cdot5^2\)
\(=\left\{100:\left[45+5\right]-32\right\}\cdot5-2\cdot25\)
\(=\left\{100:50-32\right\}\cdot5-50\)
\(=\left(-30\right)\cdot5-50\)
=-150-50
=-200
b: \(\left(25-16-3^2\right)\cdot2024\)
\(=\left(9-9\right)\cdot2024\)
\(=0\cdot2024=0\)
c: \(\left(34-42-2023\right)\cdot128^0\)
\(=34-42-2023\)
=-8-2023
=-2031
\(4^{x+2}.5.4=1280\Leftrightarrow4^{x+2}=\frac{1280}{5.4}\Leftrightarrow4^{x+2}=64\Leftrightarrow4^{x+2}=4^3\)
\(\Rightarrow x+2=3\Leftrightarrow x=3-2\Leftrightarrow x=1\)
4x+2.5.4=1280
4x+2.20=1280
4x+2 =1280:20
4x.42 =64
4x.16=64
4x =64:16
4x=4
=>x=1