Violympic toán 6
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6 tháng 3 2021 lúc 12:52
Cho A = \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+.....+\dfrac{1}{2019^2}\)
Chứng minh rằng \(\dfrac{20}{101}< A< \dfrac{1}{4}\)
Cho \(A=\dfrac{1}{3^2}+\dfrac{1}{6^2}+\dfrac{1}{9^2}+...+\dfrac{1}{9n^2}.\)
Chứng tỏ rằng
A\(< \dfrac{2}{9}\)
a) \(\dfrac{\left(3+\dfrac{1}{6}\right)-\dfrac{2}{5}}{\left(5-\dfrac{1}{6}\right)+\dfrac{7}{10}}\)
b) \(\dfrac{\left(4,08-\dfrac{2}{25}\right):\dfrac{4}{17}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}\right).2\dfrac{2}{7}}\)
c) \(\dfrac{2-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{3}{5}}{3-\dfrac{1}{5}-\dfrac{5}{3}}\)
chứng minh rằng:
\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
Chứng minh rằng: \(\dfrac{3}{2^2}+\dfrac{5}{6^2}+\dfrac{7}{12^2}+\dfrac{9}{20^2}+...+\dfrac{19}{90^2}< 1\)
Chứng tỏ rằng:
\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{2}\)
Hãy chứng tỏ rằng: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{n^2}\)<1
2. Chứng minh rằng
\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)+......+\(\dfrac{1}{2007^2}\) < \(\dfrac{1}{4}\)
Help me !!
Bài 1: Tính:
a) 10,(3) + 0,(4) - 8,(6)
b) \(\dfrac{0,8:\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}\right).2\dfrac{2}{17}}+\left(1,2.0,5\right):\dfrac{4}{5}\)