\(3+\frac{3}{2}+\frac{3}{2^2}+........+\frac{3}{2mũ}9\)
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
20 tháng 7 2019
\(\frac{\frac{2}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}+\frac{2}{13}}{\frac{3}{5}+\frac{3}{7}+\frac{3}{9}+\frac{3}{11}+\frac{3}{13}}\)+\(\frac{15151515}{45454545}\)=\(\frac{2}{3}\)(\(\frac{\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}}{\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}}\))+\(\frac{15.1010101}{45.1010101}\)
=\(\frac{2}{3}\)+\(\frac{15}{45}\)
=\(\frac{2}{3}\)+\(\frac{1}{3}\)=1
CA
1 tháng 4 2016
\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)
\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)
13 tháng 12 2018
\(a)A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}\)
\(=\frac{(23+1)\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}=\frac{47-23+24}{47-23+24}\cdot\frac{3(1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13})}{3(3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11})}\)
\(=\frac{1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13}}{3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11}}=\frac{1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11}}{3(1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11})}=\frac{1}{3}\)
\(b)\)\(\text{Đặt A = }1+2+2^2+2^3+...+2^{2012}\)
\(2A=2(1+2^2+2^3+...+2^{2012})\)
\(2A=2+2^2+2^3+...+2^{2013}\)
\(2A-A=(2+2^2+2^3+2^4+...+2^{2013})-(1+2+2^2+2^3+...+2^{2012})\)
\(\Rightarrow A=2^{2013}-1\)
\(\text{Quay lại bài toán,ta có :}\)
\(B=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2(2^{2013}-1)}=\frac{1}{2}\)
TN
0
Đặt \(A=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Rightarrow A=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
Đặt \(S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Rightarrow2S=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)
\(\Rightarrow2S-S=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)\)
\(\Rightarrow S=2-\frac{1}{2^9}\)
Mà \(A=3.S\)
\(\Rightarrow A=3.\left(2-\frac{1}{2^9}\right)\)
\(\Rightarrow A=6-\frac{3}{2^9}\)
Chúc bạn học tốt !!!