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M = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
M = \(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11}\)
M = \(\frac{2}{1}-\frac{2}{11}\)
M = \(\frac{20}{11}\)
a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
n=\(\frac{2}{3}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\times\frac{98}{99}\)
n=\(\frac{196}{297}\)
Câu \(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{2}{99.100}\)Bạn viết \(\frac{3}{99.100}=\frac{2}{99.100}\)mik sửa lại nhé.
\(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\)
\(M=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{100-99}{99.100}\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}\)
\(N=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{99-97}{97.99}\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{3}{2}.\frac{98}{99}=\frac{49}{33}\)
Ta thấy : \(\frac{297}{200}>\frac{49}{33}\Rightarrow M>N\)
a) Đặt A= \(\frac{1+2+2^2+2^3+...+2^{2009}}{1-2^{2010}}\)
Đặt S = 1 + 2 + 22 + 23 + ... + 22009
=> 2S = 2 + 22 + 23 + ... + 22010
=> 2S - S = (2 + 22 + 23 + ... + 22010) - (1 + 2 + 22 + 23 + .. + 22009)
=> S = 22010 - 1
=> S = - 1 - 22010
\(\Rightarrow A=\frac{-1-2^{2010}}{1-2^{2010}}=-1\)
Vậy \(\frac{1+2+2^2+2^3+...+2^{2009}}{1-2^{2010}}=-1\)
b) Đặt: \(A=\frac{1}{299.297}-\frac{1}{297.295}-\frac{1}{295.293}-...-\frac{1}{3.1}\)
\(\Rightarrow-2A=-\frac{2}{299.297}+\frac{2}{297.295}+\frac{2}{295.293}+...+\frac{2}{3.1}\)
\(\Rightarrow-2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{295.297}-\frac{2}{297.299}\)
\(\Rightarrow-2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{295}-\frac{1}{297}-\frac{1}{297.299}\)
\(\Rightarrow-2A=1-\frac{1}{297}-\frac{2}{88803}\)
\(\Rightarrow-2A=\frac{296}{297}-\frac{2}{88803}=\frac{88504}{88803}-\frac{2}{88803}=\frac{88502}{88803}\)
\(\Rightarrow A=\frac{88502}{88803}:\left(-2\right)=\frac{44251}{88803}\)
Vậy \(\frac{1}{299.297}-\frac{1}{297.295}-\frac{1}{295.293}-...-\frac{1}{3.1}=\frac{44251}{88803}\)
c) Đặt \(B=\frac{12}{1.3.5}+\frac{12}{3.5.7}+\frac{12}{5.7.9}+...+\frac{12}{25.27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{12}{25.27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{1}{1.3}-\frac{1}{27.29}\)
\(\Rightarrow\frac{B}{3}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow B=\frac{260}{783}.3=\frac{260}{261}\)
Vậy \(\frac{12}{1.3.5}+\frac{12}{3.5.7}+\frac{12}{5.7.9}+...+\frac{12}{25.27.29}=\frac{260}{261}\)
Duyệt mk nha!!!
\(\frac{2.2}{1.3}x\frac{3.3}{2.4}x\frac{4.4}{3.5}x\frac{5.5}{4.6}x\frac{6.6}{5.7}\)=\(2.\frac{2}{3}.\frac{3}{2}.\frac{3}{4}.\frac{4}{3}.\frac{4}{5}.\frac{5}{4}.\frac{5}{6}.\frac{6}{5}.\frac{6}{7}\)
\(=2.\frac{6}{7}=\frac{12}{7}\)
22/1.3 × 32/2.4 × 42/3.5 × 52/4.6 × 62/5.7
= 2.3.4.5.6/1.2.3.4.5 × 2.3.4.5.6/3.4.5.6.7
= 6 × 2/7
= 12/7
d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)
\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)
\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)
\(=-\frac{2018}{2018}\)
\(=-1\)
\(\frac{-2}{1.3}-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)
\(=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=-\left(1-\frac{1}{9}\right)\)
\(=-\frac{8}{9}\)