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Tớ không chắc thế này có phải tính nhanh hay ko nữa 0w0?
Tớ thấy các tử số ở đây đều chia hết cho 32 nên:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\)\(\frac{1}{32}\)
= \(\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}+\frac{1}{32}\)
= \(\frac{16+8+4+2+1+1}{32}\)
= \(\frac{32}{32}=1\)
#z
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}\)
\(=\frac{126}{128}=\frac{63}{64}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=\frac{1+1+1+1+1+1+1}{2}\)
\(=\frac{7}{2}\)
Đặt \(T=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(T=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)\)
\(\Rightarrow T=1-\frac{1}{128}=\frac{127}{128}\)
2y=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
2y-y=1-\(\frac{1}{256}\)
y=\(\frac{255}{256}\)
ta có : A=1/2+1/4+..+1/1024
=> A=1/21+1/22+..+1/210
=> A.2=(1/21+1/22+..+1/210).2
=> A.2=1+1/21+1/22+..+1/29
=> 2A-A=(1+1/21+1/22+..+1/29)-(1/21+1/22+..+1/210)
=> A=1-1/210
ta có:A=1/2+1/2^2+1/2^3+...+1/2^6+1/2^7 (1)
2A= 2.(1/2+1/2^2+1/2^3+...+1/2^6+1/2^7)
=1+1/2+1/2^2+....+1/2^6+1/2^7 (2)
lấy (2) trừ (1) vế với vế ta được:
2A-A=(1+1/2+1/2^2+....+1/2^6+1/2^7)-(1/2+1/2^2+...+1/2^6+1/2^7)
A=1-1/2^7
VẬY A=1-1/2^7
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{128}-\frac{1}{256}\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
**** cho mh nha bn . thanks
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}...-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}\)
\(A=1-0+0+0+...+0+0-\frac{1}{256}\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{14.15}+\frac{1}{15.16}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)
\(A\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{256}\right)\cdot2\)
\(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}\)
\(A\cdot2-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}=\frac{255}{256}\)
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^7}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(A=1-\frac{1}{2^8}\)
\(A=\frac{2^8-1}{2^8}\)
\(A=\frac{255}{256}\)