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\(A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)-\frac{1}{27}\)
\(=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)-\frac{1}{27}\)
\(=-\left(1-\frac{1}{27}\right)-\frac{1}{27}\)
\(=-1+\frac{1}{27}-\frac{1}{27}\)
\(=-1\)
\(A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)-\frac{1}{27}\)
\(=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)-\frac{1}{27}\)
\(=-\left(1-\frac{1}{27}\right)-\frac{1}{27}\)
\(=-1+\frac{1}{27}-\frac{1}{27}\)
\(=-1\)
A=\(\dfrac{2}{1.3}-\dfrac{2}{3.5}-\dfrac{2}{5.7}-.....-\dfrac{2}{23.25}-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{23.25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{23}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)
A=\(\dfrac{2}{3}-\dfrac{22}{75}-\dfrac{1}{27}\)
A=\(\dfrac{227}{675}\)
\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{99x101}\)
\(=1-\frac{1}{3}+\frac{1}{3}+\frac{1}{5}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}+\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
nếu ko làm đc thì mình làm cho nha
\(=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)-\frac{2}{27}\)
\(=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)-\frac{2}{27}\)
\(=-\left(1-\frac{1}{27}\right)-\frac{2}{27}\)
\(=-1+\frac{1}{27}-\frac{2}{27}\)
\(=-\frac{28}{27}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}=\frac{\left(2.3.4.5\right).\left(2.3.4.5\right)}{\left(1.2.3.4\right).\left(3.4.5.6\right)}=\frac{5.2}{1.6}=\frac{5}{3}\)
C = \(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}.\frac{56}{305}=\frac{74}{305}\)
Bài làm:
1) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\)
2) \(B=\frac{2^2.3^2.4^2.5^2}{1.2.3^2.4^2.5.6}=\frac{2.5}{6}=\frac{5}{3}\)
3) \(C=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(C=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{61-59}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)