8^x : 2^x = 16 ^2011
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ta có :
3n+3+3n+1+2n+2+2n+1
= 3n.(33+3)+2n.(22+2)
= 3n.30 + 2n.6 ⋮ 6
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
= \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)\)
= \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
Vậy \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\) = \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\) hay
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49}-\dfrac{1}{50}\) = \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
( x - 2/2 )^2 = 1/4
=> ( x - 2/2 )^2 = 1/2^2
=> x -2/2 = 1/2
=> x = 1/2 + 2/2
=> x = 3/2
Vậy x = 3/2
\(\dfrac{8^x}{2^x}=16^{2011}\)
\(\Leftrightarrow4^x=4^{4022}\)
\(\Rightarrow x=4022\)