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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=\(\dfrac{227}{675}\)
\(=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)-\frac{2}{27}\)
\(=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)-\frac{2}{27}\)
\(=-\left(1-\frac{1}{27}\right)-\frac{2}{27}\)
\(=-1+\frac{1}{27}-\frac{2}{27}\)
\(=-\frac{28}{27}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2025\cdot2016}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2025}-\dfrac{1}{2026}\\ =\dfrac{1}{1}-\dfrac{1}{2026}=\dfrac{2025}{2026}\\ B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =\dfrac{1}{1}-\dfrac{1}{101}=\dfrac{100}{101}\\ C=\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{99\cdot101}\\ =\dfrac{5}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{101}\right)=\dfrac{5}{2}\cdot\dfrac{100}{101}=\dfrac{250}{101}\)
\(D=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\\ =\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}=\dfrac{99}{100}\)
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Là -1.Đúng 100%