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a) \(\frac{1}{10}-\frac{1}{40}-\frac{1}{88}-\frac{1}{154}-\frac{1}{238}-\frac{1}{340}\)
\(=\frac{1}{10}-\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(=\frac{1}{10}-\frac{1}{3}.\frac{3}{20}\)
\(=\frac{1}{10}-\frac{1}{20}=\frac{2}{20}-\frac{1}{20}=\frac{1}{20}\)
Đặt A= \(\frac{1}{10}-\frac{1}{40}-..-\frac{1}{340}\)
A=\(\frac{1}{10}-\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\right)\)
3A= \(\frac{1}{10}-(\frac{3}{5.8}+...+\frac{3}{17.20})\)
3A=\(\frac{1}{10}-\left(\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
3A=\(\frac{1}{10}-\left(\frac{1}{5}-\frac{1}{20}\right)\)
3A=\(\frac{1}{10}-\frac{3}{20}\)
3A=\(-\frac{1}{20}\)
A=\(-\frac{1}{60}\)
ttiikk nha bạn
\(=\frac{1}{2.5}-\frac{1}{5.8}-...-\frac{1}{17.20}.\)
\(=\frac{1}{10}-\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right).\)
\(=\frac{1}{10}-\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right).\)
\(=\frac{1}{10}-\frac{1}{3}\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(=\frac{1}{10}-\frac{1}{3}.\frac{3}{20}\)
\(=\frac{1}{20}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
\(\Rightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)
\(\Rightarrow3x=\frac{4949}{19800}-\frac{451}{8120}\)
\(\Rightarrow x=\left(\frac{4949}{19800}-\frac{451}{8120}\right):3\)
\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)
\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3x+2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)
\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)
\(\frac{1}{2}-\frac{1}{3x+5}=\frac{12}{25}\)
\(\frac{1}{3x+5}=\frac{1}{50}\)
=> 3x+5 = 50
3x = 45
x = 15
Ta có : \(\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)\left(\frac{1}{49}-\frac{1}{4^2}\right).......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)
\(=\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)......\left(\frac{1}{49}-\frac{1}{7^2}\right)......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)
\(=\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)......0......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)
\(=0\)
1 Tính :
a) \(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{n}\)
\(=\frac{1}{n}\)
b) \(B=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(=\frac{4}{1.5}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\frac{1}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
c) \(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow C=1-B\left(1\right)\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
Lấy 2B trừ B ta có :
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(B=1-\frac{1}{2^{10}}\left(2\right)\)
Thay (2) vào (1) ta có :
\(C=1-\left(1-\frac{1}{10}\right)\)
\(=1-1+\frac{1}{10}\)
\(=\frac{1}{10}\)
Vậy \(C=\frac{1}{10}\)
Ta có:
\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
đặt \(A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)
\(A-\frac{1}{2}A=\frac{1}{2}A\Rightarrow A=\frac{1-\frac{1}{2^{11}}}{\frac{1}{2}}=2-\frac{1}{2^{10}}\)
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=-1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)
Vậy, \(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-1-A=-1-\frac{1023}{1024}=-\frac{2047}{1024}\)