Tính nhanh :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{210}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+.....+\frac{1}{1+2+3+....+50}\)
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
17 tháng 6 2018
a,Ta có \(\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{1-\frac{2}{3}-\frac{1}{2}}-\frac{\frac{3}{5}-\frac{3}{7}-\frac{3}{11}}{\frac{6}{5}-\frac{6}{7}-\frac{6}{11}}\)
\(=\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{2.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)}-\frac{3.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}{6.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}\)
=\(\frac{1}{2}-\frac{3}{6}=\frac{1}{2}-\frac{1}{2}=0\)
Vậy giá trị biểu thức bằng 0
b, Mình không hiểu cho lắm ạ , nếu ko phiền xin xem lại đầu bài ạ
26 tháng 6 2017
Đây mà toán lớp 5 à.
Áp dụng công thức
\(\frac{1}{1+2+...+n}=\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{2}{n\left(n+1\right)}\) ta được
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+....+50}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=\frac{49}{51}\)
26 tháng 6 2017
Ta có : \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3+......+50}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+......+\frac{1}{\frac{50.51}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{50.51}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{50.51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{51}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{51}\)
\(=1-\frac{2}{51}=\frac{49}{51}\)
AH
15 tháng 8 2016
- \(A=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)
A= 1/3 + 1/4-1/4+1/5-1/5+1/6-1/6+1/7-1/7+1/8-1/8+1/9-1/9+1/10
A=1/3+1/10
A=13/30
VT
15 tháng 8 2016
a,\(A=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{8.9}-\frac{1}{9.10}\)
\(=\frac{1}{12}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\frac{1}{4}+\frac{1}{10}=\frac{5}{60}-\frac{15}{60}+\frac{6}{60}=\frac{-1}{15}\)
Vậy \(A=\frac{-1}{15}\)
Đặt \(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{210}\)
\(\frac{1}{2}B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\)
\(\frac{1}{2}B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow B=\frac{\frac{1}{2}-\frac{1}{21}}{\frac{1}{2}}=\frac{19}{21}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+50}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{\left(1+50\right).50}{2}}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{1275}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{2550}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{50.51}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)