Question

cần giúp gấp ạ, em cảm ưn ạ

Answer 1

a/\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{100}{101}\)
\(=\dfrac{50}{101}\)
b/\(B=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}...+\dfrac{4}{49.51}\)
\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=2\left(1-\dfrac{1}{51}\right)\)
\(=2\cdot\dfrac{50}{51}\)
\(=\dfrac{100}{51}\)
c/\(C=\dfrac{6}{3.5}+\dfrac{6}{5.7}+\dfrac{6}{7.9}+...+\dfrac{6}{99.101}\)
\(=3\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)
\(=3\cdot\dfrac{98}{303}\)
\(=\dfrac{98}{101}\)
d/\(D=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{1023}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{31.33}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{31.33}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{33}\)
\(=\dfrac{16}{33}\)
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