Đặt tổng trên là: \(A\)

\(A=\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\)

\(\Rightarrow A.4=6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\)

\(\Rightarrow A.4-A=\left(6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\right)-\left(\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\right)\)

\(\Rightarrow A.3=6-\dfrac{3}{512}=\dfrac{3069}{512}\)

\(\Rightarrow A=\dfrac{3069}{512}:3=\dfrac{1023}{512}\)

3/512 là gì vậy bn?

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

\(=3.\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)

\(=3.A\)với \(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\)

\(\Rightarrow2^2A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}\right)\)

\(\Rightarrow2^2A-A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)

\(\Rightarrow4A-A=2-\frac{1}{2^9}\)

\(\Rightarrow3A=2-\frac{1}{512}=\frac{1023}{512}\Rightarrow A=\frac{1023}{512}:3\)

\(\Rightarrow\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=3.\left(\frac{1023}{512}:3\right)=\frac{1023}{512}\)

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

\(=\left(\frac{12}{8}+\frac{3}{8}\right)+\left(\frac{12}{128}+\frac{3}{128}\right)+\frac{3}{512}\)

\(=\frac{15}{8}+\frac{15}{128}+\frac{3}{512}\)

\(=\frac{240}{128}+\frac{15}{128}+\frac{3}{512}\)

\(=\frac{255}{128}+\frac{3}{512}\)

\(=\frac{1020}{512}+\frac{3}{512}\)

\(=\frac{1023}{512}\)

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
                                                        \(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\frac{3}{8}+\frac{3}{16}-\frac{3}{16}+\frac{3}{32}\)
                                                         \(=3+\frac{3}{32}=\frac{3.32}{32}+\frac{3}{32}=\frac{96+3}{32}=\frac{99}{32}\)

0,856m=?m

\(...=\dfrac{3}{2}x\left(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\right)\)

\(=\dfrac{3}{2}x\left(\dfrac{256}{256}+\dfrac{64}{256}+\dfrac{16}{256}+\dfrac{4}{256}+\dfrac{1}{256}\right)\)

\(=\dfrac{3}{2}x\dfrac{341}{256}=\dfrac{1023}{512}\)

(3/2+3/8) +( 3/4 =3/16) +( 3/32+ 3/128)+(3,64+3,256)+3/512

= 4+6/2+6/16+6/896+3/512

=10/2+13/056+3/512

=23/256+3/512

=26/768

\(P+\frac{1}{512}=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{4}{512}=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{4}{128}=\)

\(=\frac{3}{2}+\frac{3}{8}+\frac{4}{32}=\frac{3}{2}+\frac{4}{8}=\frac{4}{2}=2\)

\(\Rightarrow P=2-\frac{1}{512}=\frac{1023}{512}\)

\(P=\frac{3}{2}+\frac{3}{8}+...+\frac{3}{512}\)

\(=3.\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)

\(4P=3\left(\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{11}}\right)\)

\(4P-P=3\left(\frac{1}{2}-\frac{1}{2^{11}}\right)\)

\(3P=3\left(\frac{1}{2}-\frac{1}{2^{11}}\right)\)

\(P=\frac{1}{2}-\frac{1}{2^{11}}=\frac{2^{10}-1}{2^{11}}\)

Đinh Thùy LinhTính nhanh mà bạn cách này mk nhanh hơn nhiều :D