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\(\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{512}+\frac{1}{2048}\)
= \(\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}+\frac{1}{2^{11}}\)
= \(\frac{341}{2048}\)
Đặt A=1+2+4+...+128=1+2+22+...+27
=>2A=2+22+23+...+28
=>2A+1=1+2+22+...+27+28=A+28
=>2A-A=28-1
A=256-1
A=255
voloc
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{64}-\frac{1}{128}\)
\(A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}.\)
\(=\frac{1}{1.2}+\frac{1}{2.2}+...+\frac{1}{64.2}\)
Rồi còn lại bạn tự giải nha
CHúc bạn học tốt
gọi biểu thức là A
A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10
=>2A=1+1/2+1/2^2+...+1/2^9
=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10
Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048
A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11
2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10
2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)
A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11
A= 1- 1/2^11
A= 2047/ 2048
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+...+\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(A=1-\frac{1}{2048}\)
\(\Rightarrow\)\(A=\frac{2047}{2048}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3B-B=1-\frac{1}{2187}\)
\(2B=\frac{2186}{2187}\)
\(\Rightarrow B=\frac{2186}{4374}=\frac{1093}{2187}\)
Đặt \(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..-\frac{1}{2048}\)
\(\Rightarrow A=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-..-\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(\Rightarrow A=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-..-\frac{1}{1024}+\frac{1}{2018}\)
\(\Rightarrow A+\frac{1}{2018}\)
1-1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512-1/1024-1/2048 =0.00048828125
Dễ thôi mà
\(\frac{25\times4-0,5\times40\times5\times0,2\times20\times0,25}{1+2+4+8+...+128+256}\)\(=\frac{100-\left(0,5\times20\right)\times\left(40\times0,25\right)\times\left(5\times0,2\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100-10\times10\times1}{1+2+4+8+...+128+256}\)\(=\frac{100-100}{1+2+4+8+...+128+256}\)\(=\frac{0}{1+2+4+8+...+128+256}\)= 0
Chúc e học tốt!
\(=\frac{1}{1x2}+\frac{1}{2x4}+\frac{1}{4x8} +\frac{1}{8x16}+\frac{1}{16x32}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
nè ..!
\(\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{512}\)
\(=\frac{256+64+16+4+1}{512}\)
\(=\frac{341}{512}\)
Ta có: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
= \(\frac{1}{2}+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+....+\left(\frac{1}{64}-\frac{1}{128}\right)\)
=\(\frac{1}{2}+\frac{1}{2}-\frac{1}{128}\)
\(=1-\frac{1}{128}=\frac{127}{128}\)
Đặt \(A=\frac12+\frac18+\cdots+\frac{1}{2048}+\frac{1}{8192}\)
=>\(4A=2+\frac12+\cdots+\frac{1}{512}+\frac{1}{2048}\)
=>\(4A-A=2+\frac12+_{\ldots}+\frac{1}{512}+\frac{1}{2048}-\frac12-\frac18-\cdots-\frac{1}{8192}\)
=>\(3A=2-\frac{1}{8192}=\frac{16383}{8192}\)
=>\(A=\frac{16383}{8192}:3=\frac{5461}{8192}\)
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