(1+1/2) x (1+1/3) x ..... x ( 1 x 1/2005)
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a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110
a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5
A = ( 1 - \(\dfrac{1}{5}\)).(1 - \(\dfrac{2}{5}\)).(1 - \(\dfrac{3}{5}\))...(1 - \(\dfrac{2005}{5}\))
A = (1- \(\dfrac{1}{5}\)).(1 - \(\dfrac{2}{5}\)).(1-\(\dfrac{3}{5}\)).(1 - \(\dfrac{4}{5}\)).(1 - \(\dfrac{5}{5}\))....(1 - \(\dfrac{2005}{5}\))
A = (1 - \(\dfrac{1}{5}\)).(1- \(\dfrac{2}{5}\)).(1- \(\dfrac{3}{5}\)).(1- \(\dfrac{4}{5}\)).(1-1)....(1- \(\dfrac{2005}{5}\))
A = (1- \(\dfrac{1}{5}\)).....0...(1- \(\dfrac{2005}{5}\))
A =0
a)\(=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times...\times\left(\dfrac{2005}{2005}+\dfrac{1}{2005}\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times...\times\dfrac{2006}{2005}=\dfrac{2006}{2}=1003\)
b)\(=\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\times\dfrac{1}{2}=\dfrac{3}{3}\times\dfrac{1}{2}=\dfrac{1}{2}\)
\(\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times..........\times\left(1+\frac{1}{2005}\right)\)
=\(\frac{3}{2}\times\frac{4}{3}\times............\times\frac{2006}{2005}\)
=\(\frac{2006}{2}=1003\)
3 4 2006
_x _x ......x____=2006/2=1003
2 3 2005