1.Cho G= \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2^{1024}}\right)\)và H=\(\frac{1}{2^{2047}}\)Tính G+H
cho \(G=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2^{1024}}\right)\)và \(H=\frac{1}{2^{2047}}\). Tính G+H
Cho \(G=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2^{1024}}\right)\)và \(H=\frac{1}{2^{2047}}\). Tính G+H
cho \(G=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2^{1024}}\right)\)và \(H=\frac{1}{2^{2047}}\). Tính G+H
Cho \(G=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{16}\right)\left(1+\frac{1}{256}\right)...\left(1+\frac{1}{2^{1024}}\right)\)và \(H=\frac{1}{2^{2047}}\)
Tính \(G+H\)
CHO \(G=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{16}\right)+\left(1+\frac{1}{256}\right)+...+\left(1+\frac{1}{2^{1020}}\right)\)
\(H=\frac{1}{2^{2047}}\) .Tính\(G+H\)
cho G = \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)+\left(1+\dfrac{1}{256}\right)......\left(1+\dfrac{1}{2^{1024}}\right)\)và H = \(\dfrac{1}{2^{2047}}\)
Tính G + H
BÀI 2 : Tính
a) \(\left(-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{h}\right)\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right).....\left(1+\frac{1}{h}\right)\)
c) \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\)
\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+....+\left(1-\frac{1}{512}\right)+\left(1-\frac{1}{1024}\right)\)
Tính
\(F=\left[\left(0.1\right)^2\right]^0+\left[\left(\frac{1}{7}\right)^{-1}\right]^2.\frac{1}{19}\left[\left(2^2\right)^3:2^5\right]\)
\(G=\left(xy\right)^{-2}.\left[\frac{1}{2}y:x\right]^3\)
\(H=\frac{1}{1-\frac{1}{1-2^{-1}}}+\frac{1}{1+\frac{1}{1+2^{-1}}}\)