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\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+...+\dfrac{4056194}{4056195}+\left(-1006\right)\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{4056195}\right)+\left(-1006\right)\\ =\left[1+1+1+...+1+\left(-1006\right)\right]-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{4056195}\right)\\ =1-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{4056195}\right)\\ =1-\dfrac{1}{2}\cdot\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{4056195}\right)\\ =1-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2013\cdot2015}\right)\\ =1-\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\\ =1-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2015}\right)\\ =1-\dfrac{1}{2}\cdot\dfrac{2014}{2015}\\ =1-\dfrac{1007}{2015}\\ =\dfrac{1008}{2015}\)
\(A=\dfrac{6}{3}+\dfrac{6}{15}+\dfrac{6}{35}+...+\dfrac{6}{9999}\)
\(=3\left(\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\right)\)
\(=3\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=3\left(1-\dfrac{1}{101}\right)=3\cdot\dfrac{100}{101}=\dfrac{300}{101}\)
\(A=\dfrac{-1}{3}+\dfrac{-1}{15}+\dfrac{-1}{35}+...+\dfrac{-1}{9999}\)
\(\Rightarrow-A=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\)
\(-A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(-2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(-2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(-2A=1-\dfrac{1}{101}\)
\(-2A=\dfrac{100}{101}\)
\(-A=\dfrac{100}{101}:2\)
\(-A=\dfrac{50}{101}\)
\(\Rightarrow A=\dfrac{-50}{101}\)
Chúc bạn học tốt!
\(A=\dfrac{-1}{3}+\dfrac{-1}{15}+\dfrac{-1}{35}+...+\dfrac{-1}{9999}\)
\(A=-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\)
Đặt \(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+....+\dfrac{1}{9999}\)
\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(2B=1-\dfrac{1}{101}=\dfrac{100}{101}\)
\(B=\dfrac{100}{101}:2=\dfrac{50}{101}\)
\(\Rightarrow A=-B=-\dfrac{50}{101}\)
\(M=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+...+1-\frac{1}{9999}\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)
\(M=50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(M=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(M=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{5050-100}{101}=\frac{4950}{101}\)
\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+...+\dfrac{9997}{9999}\)
\(=1-\dfrac{2}{3}+1-\dfrac{2}{15}+1-\dfrac{2}{35}+...+1-\dfrac{2}{9999}\)
\(=\left(1+1+1+...+1\right)-\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\)
\(=50-1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=50-\left(1-\dfrac{1}{101}\right)=50-\dfrac{100}{101}\)
\(=\dfrac{4950}{101}\)
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
a) Sửa tí: \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
\(\Rightarrow2A=2.\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(\Rightarrow2A-A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2006}}\right)\)
\(\Rightarrow A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2006}}\)
\(\Rightarrow A=2-\dfrac{1}{2^{2006}}\)
b) Đặt \(A=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{1}{50.61}\)
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(A=\dfrac{1}{5}-\dfrac{1}{61}\)
\(A=\dfrac{56}{305}\)
c) Đặt \(A=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)
\(A=\dfrac{7}{2}.2.\left(\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{9999}\right)\)
\(A=\dfrac{7}{2}.\left(1-\dfrac{1}{101}\right)\)
\(A=\dfrac{7}{2}.\dfrac{100}{101}\)
\(A=\dfrac{256}{101}\)
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\dfrac{96}{505}\)
\(\Rightarrow G=\dfrac{64}{505}\)
\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+...+\dfrac{9998}{9999}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{9999}\right)\\ =\left(1+1+1+...+1\right)\left(\text{có 50 số 1}\right)-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\\ =50\cdot1-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\right)\\ =50-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =50-\left(1-\dfrac{1}{101}\right)\\ =50-1+\dfrac{1}{101}\\ =49+\dfrac{1}{101}\\ =\dfrac{4949+1}{101}\\ =\dfrac{4950}{101}\)