Giải:
\(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)
\(A=\left(\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\right).\dfrac{1}{2}\)
\(A=\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\right).\dfrac{1}{2}\)
\(A=\left(1-\dfrac{1}{41}\right).\dfrac{1}{2}\)
\(A=\dfrac{40}{41}.\dfrac{1}{2}\)
\(A=\dfrac{20}{41}\)
\(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)
\(B=\left(\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\right).\dfrac{1}{3}\)
\(B=\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}\right).\dfrac{1}{3}\)
\(B=\left(1-\dfrac{1}{31}\right).\dfrac{1}{3}\)
\(B=\dfrac{30}{31}.\dfrac{1}{3}\)
\(B=\dfrac{10}{31}\)
Vì \(\dfrac{20}{41}>\dfrac{10}{31}\) nên \(A>B\)
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