tính nhanh hộ mình câu này với : (có kèm bài làm nha )
\(\frac{3}{2}+\frac{ 3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(5+\frac{5}{3}+\frac{5}{9}+\frac{5}{27}+\frac{5}{81}\)
\(\frac{4}{3}+\frac{1}{3}+\frac{1}{12}+\frac{1}{48}+\frac{1}{192}\)
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\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(=3.\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)
\(=3.A\)với \(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\)
\(\Rightarrow2^2A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}\right)\)
\(\Rightarrow2^2A-A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)
\(\Rightarrow4A-A=2-\frac{1}{2^9}\)
\(\Rightarrow3A=2-\frac{1}{512}=\frac{1023}{512}\Rightarrow A=\frac{1023}{512}:3\)
\(\Rightarrow\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=3.\left(\frac{1023}{512}:3\right)=\frac{1023}{512}\)
Sao mà mình hỏi bài này từ lâu lắm rồi mà vẫn chưa có bạn nào trả lời nhỉ?
A) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
2A= \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
2A-A = \(1-\dfrac{1}{32}\)
A= \(\dfrac{31}{32}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)
=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)
a)=768/512+192/512+48/512+12/512+3/512
=768+192+48+12+3/512
=1023/512
b)=405/81+135/81+45/81+15/81+5/81
=405+135+45+15+5/81
=595/81
c)=256/192+64/192+16/192+4/192+1/192
=256+64+16+4+1/192
=341/192