Tính tổng
A = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+8+9}\)
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Những câu hỏi liên quan
30 tháng 10 2016
tử số cộng thêm 1 vào mỗi phân số ở tử
là ra ngay nha bạn
k mình nha
NH
2
S
12 tháng 7 2018
\(\frac{\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=\frac{\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\)
HN
12 tháng 7 2018
Đặt S = \(\frac{\frac{9}{1}+\frac{8}{2}+...........+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{10}}\)
Đặt A là tử số, B là mẫu số
Xét A:
\(A=\frac{9}{1}+\frac{8}{2}+............+\frac{1}{9}\)
\(A=\left(9-1-1-......-1\right)+\left(\frac{8}{2}+1\right)+.........+\left(\frac{1}{9}+1\right)\)
\(A=1+\frac{10}{2}+.......+\frac{10}{9}\)
\(A=\frac{10}{1}+\frac{10}{2}+........+\frac{10}{9}\)
\(A=10\left(\frac{1}{2}+........+\frac{1}{9}\right)\)
Thay vào S ta có:
\(S=\frac{10\left(\frac{1}{2}+......+\frac{1}{9}\right)}{\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{10}}=10\)
Vậy S = 10
TN
21 tháng 8 2015
\(A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^8}+\frac{1}{3^9}\)
\(3A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}+\frac{1}{3^8}\)
\(3A-A=\frac{1}{3}-\frac{1}{3^9}\)
\(2A=\frac{1}{3}.\left(1-\frac{1}{3^8}\right)\)
\(A=\frac{1}{6}.\left(1-\frac{1}{3^8}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n-1}}+\frac{1}{2^n}\)
\(\frac{1}{2}B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}+\frac{1}{2^{n+1}}\)
\(B-\frac{1}{2}B=1-\frac{1}{2^{n+1}}\)
\(\frac{1}{2}B=1-\frac{1}{2^{n+1}}\)
\(B=2-\frac{2}{2^n.2}=2-\frac{1}{2^n}\)
21 tháng 6 2015
a) \(\frac{\left(-1\right)}{4}^2+\frac{3}{8}.\left(\frac{-1}{6}\right)-\frac{3}{16}:\left(\frac{-1}{2}\right)=\left(\frac{-1}{4}\right)^2+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\left(\frac{1}{16}\right)+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\frac{5}{272}-\left(\frac{-3}{8}\right)=\frac{107}{272}\)
HT
15 tháng 8 2015
A = \(\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+...+\frac{1}{9}.\frac{1}{10}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
A = \(1-\frac{1}{10}\)
A = \(\frac{9}{10}\)
15 tháng 8 2015
1/2=1-1/2 ; 1/2.1/3=1/2-1/3 ; 1/3.1/4=1/3-1/4...v...v
Vậy A bằng: 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.............+1/8-1/9+1/9-1/10
=1-1/10=9/10
\(A=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+9}=\frac{1}{1.2:2}+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{9.10:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{9.10}=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)=2\left(1-\frac{1}{10}\right)\)
\(=2.\frac{9}{10}=\frac{9}{5}\)
\(A=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+8+9}\)
=>\(A=\frac{2}{2}+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{9.10:2}\)
=>\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{9.10}\)
=>\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
=>\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=>\(A=2.\left(1-\frac{1}{10}\right)\)
=>\(A=2.\frac{9}{10}\)
=>\(A=\frac{9}{5}\)