chứng minh 1phần 2 - 1 phần 4 + 1 phần 8 - 1 phầm 16 + 1 phần 32 - 1 phần 64 + 1 phần 128 - 1 phần 256 < 1 phần 3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
16 tháng 5 2021
7/48 - (1/2 x 2 + 1/6 x 4 + 1/8 x 5 + 1/12 x 7 + 1/14 x 8) : x = 0
7/48 - (1 + 2/3 + 5/8 + 7/12 + 4/7) : x = 0 (đã rút gọn)
7/48 - (336/336 + 224/336 + 210/336 + 196/336 + 192/336) : x = 0 (quy đồng)
7/48 - 193/56 : x = 0
193/56 : x = 0 + 7/48
193/56 : x = 7/48
x = 193/56 : 7/48
x = 1158/49
NT
4
23 tháng 6 2016
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}.\)
\(A+\frac{1}{64}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{64}\)
\(A+\frac{1}{64}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{32}\)
\(A+\frac{1}{64}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}=...=\frac{1}{2}\)
\(A=\frac{1}{2}-\frac{1}{64}=\frac{31}{64}.\)
23 tháng 6 2016
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{16}-\frac{1}{32}\right)+\left(\frac{1}{32}-\frac{1}{64}\right)\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(=\frac{1}{2}-\frac{1}{64}=\frac{31}{64}\)
C
25 tháng 1 2018
Gọi biểu thức trên là \(A\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}\)
\(2A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}\right)\times2\)
\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+\frac{1}{16}\times2+...+\frac{1}{512}\times2\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
NT
0
15 tháng 7 2018
1/2 + 1/4 +1/8 + 1/16 + 1/32
= 16/32 + 3/32 + 4/32 + 2/32 + 1/32
=26/32 =13/16
27 tháng 7 2021
a) \(27^{64}:81^{20}=3^{192}:3^{80}=3^{112}\)
b) \(\left(\dfrac{1}{8}\right)^{20}:\left(\dfrac{1}{16}\right)^9=\left(\dfrac{1}{2}\right)^{60}:\left(\dfrac{1}{2}\right)^{36}=\left(\dfrac{1}{2}\right)^{24}\)
c) \(\dfrac{1}{3}:\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{5}{3}-\dfrac{1}{6}=\dfrac{10}{6}-\dfrac{1}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)
AT
25 tháng 3 2017
Ta có:
\(A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)
\(A>\dfrac{1}{40}.10+\dfrac{1}{50}.10+\dfrac{1}{60}.10=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{3}{5}\)
Vậy \(A>\dfrac{3}{5}\)
Ta có:
\(A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)\(A< \dfrac{1}{31}.10+\dfrac{1}{41}.10+\dfrac{1}{51}.10< \dfrac{4}{5}\)
Vậy \(A< \dfrac{4}{5}\)
Do đó: \(\dfrac{3}{5}< A< \dfrac{4}{5}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}+\frac{1}{128}-\frac{1}{256}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}\)
\(A+2A=\left(\frac{1}{2}-\frac{1}{4}+...-\frac{1}{256}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-...-\frac{1}{128}\right)\)
\(3A=1-\frac{1}{256}< 1\)
\(\Rightarrow A< \frac{1}{3}\).