Cho \(A=1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{10}-...-\dfrac{1}{512}.CM:A< 0,002\)
Những câu hỏi liên quan
A=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+....+\(\dfrac{1}{512}\)+\(\dfrac{1}{1024}\)
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)
Đúng 1
Bình luận (0)
18 tháng 3 2023 lúc 19:38
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
tính nhanh \(\dfrac{1}{2}\) +\(\dfrac{1}{4}\) +\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\) +.....+\(\dfrac{1}{512}\) +\(\dfrac{1}{1024}\)
Đặt A=1/2+1/4+1/8+..+1/1024
Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)
Ax2-A=(1+1/2+1/4+1/8+..+1/512)-(1/2+1/4+1/8+..+1/1024)
<=>A=1-1/1024
<=>A=1023/1024
Vậy biểu thức đã cho = 1023/1024
Đúng 0
Bình luận (0)
S=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{512}\)+\(\dfrac{1}{1024}\)
Câu 2 : Tính nhanh:
\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+...+\(\dfrac{1}{256}\)+\(\dfrac{1}{512}\)
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{256}-\dfrac{1}{512}\)
\(\Rightarrow A=1-\dfrac{1}{512}=\dfrac{511}{512}\)
Đúng 0
Bình luận (0)
Đặt ⇒2A=1+12+14+...+1128+1256⇒2A=1+12+14+...+1128+1256
⇒A=1−1512=511512
Đúng 0
Bình luận (0)
(2 + 4 + 6 + ... + 98) x (\(\dfrac{3}{4}\) x 8 - 15 x 0,4) x (\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{512}\) )
Giúp mình vớiii
\(=\left(2+4+6+...+98\right)\left(6-6\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)\)
=0
Đúng 5
Bình luận (0)
= ( 2+4+6+...+98 ) ( 6- 6) ( 1/2+1/4 + .......+ 1/ 512 )
= 0
Chúc bạn học tốt
Đúng 1
Bình luận (0)
= (2 + 4 + 6 + ... + 98) x ( 6 - 6 ) x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{512}\))
= (2 + 4 + 6 + ... + 98) x 0 x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{512}\))
= 0
*****Vì số nào nhân với 0 cũng bằng 0
Đúng 0
Bình luận (0)
Cho A= \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+\dfrac{1}{10^2}+...+\dfrac{1}{160^2}\)
Chứng minh: \(\dfrac{1}{8}< A< \dfrac{3}{16}\)
Cho biểu thức A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)+\(\dfrac{1}{8^2}\)+\(\dfrac{1}{9^2}\)+\(\dfrac{1}{10^2}\)
Chứng minh rằng A<1
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2\cdot2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3\cdot3}< \dfrac{1}{2\cdot3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4\cdot4}< \dfrac{1}{3\cdot4}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9\cdot9}< \dfrac{1}{8\cdot9}\)
\(\dfrac{1}{10^2}=\dfrac{1}{10\cdot10}< \dfrac{1}{9\cdot10}\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A< 1-\dfrac{1}{10}\)
\(\Rightarrow A< \dfrac{9}{10}\)
\(\Rightarrow A< 1\) (vì: \(\dfrac{9}{10}< 1\))
Đúng 2
Bình luận (0)
cho biểu thức:A=\(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...\dfrac{ }{ }\)\(\dfrac{1}{40}\)
CM:A>\(\dfrac{1}{2}\)
người nào giải đầu tiên mà nhanh mik sẽ tick
ta có A = 1/21 + 1/22 + 1/23 + 1/24 + ... + 1/40 > 1/40 + 1/40 +....+ 1/40 ( có 20 số hạng 1/40)
= 20/40
=1/2
=) A> 1/2 (1)
ta lại có A = 1/21 + 1/22 + 1/23 + 1/24 + ... + 1/40 < 1/20 + 1/20 +...+ 1/20 ( có 20 số hạng 1/20)
=20/20
=1
=) A <1 (2)
từ (1), (2) = 1/2 <A<1
Đúng 4
Bình luận (1)