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\(LINH=\dfrac{3}{1^2.2^2}+\dfrac{7}{3^2.4^2}+\dfrac{11}{5^2.6^2}+\dfrac{15}{7^2.8^2}+\dfrac{19}{9^2.10^2}\)
\(LINH=\dfrac{1+2}{1^2.2^2}+\dfrac{3+4}{3^2.4^2}+\dfrac{5+6}{5^2.6^2}+\dfrac{7+8}{7^2.8^2}+\dfrac{9+10}{9^2.10^2}\)
\(LINH=\dfrac{1}{1^2.2^2}+\dfrac{2}{1^2.2^2}+\dfrac{3}{3^2.4^2}+\dfrac{4}{3^2.4^2}+\dfrac{5}{5^2.6^2}+\dfrac{6}{5^2.6^2}+\dfrac{7}{7^2.8^2}+\dfrac{8}{7^2.8^2}+\dfrac{9}{9^2.10^2}+\dfrac{10}{9^2.10^2}\)
\(LINH=\dfrac{1}{1.2^2}+\dfrac{1}{1^2.2}+\dfrac{1}{3.4^2}+\dfrac{1}{3^2.4}+\dfrac{1}{5.6^2}+\dfrac{1}{5^2.6}+\dfrac{1}{7.8^2}+\dfrac{1}{7^2.8}+\dfrac{1}{9.10^2}+\dfrac{1}{9^2.10}\)\(LINH=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+\dfrac{1}{180}+\dfrac{1}{150}+\dfrac{1}{448}+\dfrac{1}{392}+\dfrac{1}{900}+\dfrac{1}{810}\)Vì:
\(\left\{{}\begin{matrix}\dfrac{1}{48}< \dfrac{1}{32}\\\dfrac{1}{36}< \dfrac{1}{32}\\...............\\\dfrac{1}{810}< \dfrac{1}{32}\end{matrix}\right.\)
Nên:
\(\dfrac{1}{48}+\dfrac{1}{36}+.....+\dfrac{1}{810}< \dfrac{1}{32}+\dfrac{1}{32}+....+\dfrac{1}{32}\)
\(\Rightarrow\dfrac{1}{48}+\dfrac{1}{36}+....+\dfrac{1}{810}< \dfrac{1}{32}.8=\dfrac{1}{4}\)
Nên:
\(LINH=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+....+\dfrac{1}{810}< \dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{4}=1\)
Nên \(LINH< 1\left(đpcm\right)\)
\(D=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(D=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{4^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{10^2}\right)\)
\(D=\dfrac{1}{1}-\dfrac{1}{10^2}\)
\(D=1-\dfrac{1}{100}< 1\)
Vậy \(D< 1\left(đpcm\right)\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^2}-\dfrac{1}{4^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{100^2}\right)\)
\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{19}{81.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Ta có:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
= \(\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
= \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
= \(1-\dfrac{1}{10^2}\)
Mà \(1-\dfrac{1}{10^2}< 1\) nên:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) < 1 (đpcm).
Chứng minh:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}\)
\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{10^2}\right)\)
=\(\dfrac{1}{1}-\dfrac{1}{10^2}\)
\(=1-\dfrac{1}{100}\)
Mà \(1-\dfrac{1}{100}< 1\)
\(\Rightarrow\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}< 1\) (đpcm)
\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(A=\dfrac{2^2}{1^2.2^2}-\dfrac{1^2}{1^2.2^2}+\dfrac{3^2}{2^2.3^2}-\dfrac{2^2}{2^2.3^2}+...+\dfrac{10^2}{9^2.10^2}-\dfrac{9^2}{9^2.10^2}\)\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
Ta có:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}\)
\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{10^2}\right)\)
\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=\dfrac{1}{1^2}-\dfrac{1}{10^2}\)
\(=1-\dfrac{1}{100}\)
Vì \(1-\dfrac{1}{100}< 1\)
Nên \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}< 1\) (Đpcm)
\(vt:\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2+10^2}\)
=\(\dfrac{1}{1}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+..+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
=\(\dfrac{1}{1}-\dfrac{1}{10^2}\)
=>A<1
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\)
\(A=\dfrac{3}{1^2.2^2}+\dfrac{7}{3^2.4^2}+\dfrac{11}{5^2.6^2}+\dfrac{15}{7^2.8^2}+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{1+2}{1^2.2^2}+\dfrac{3+4}{3^2.4^2}+\dfrac{5+6}{5^2.6^2}+\dfrac{7+8}{7^2.8^2}+\dfrac{9+10}{9^2.10^2}\)
\(A=\dfrac{1}{1.2^2}+\dfrac{1}{1^2.2}+\dfrac{1}{3.4^2}+\dfrac{1}{3^2.4}+\dfrac{1}{5.6^2}+\dfrac{1}{5^2.6}+...+\dfrac{1}{9^2.10}\)
\(A=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+\dfrac{1}{180}+\dfrac{1}{150}+....+\dfrac{1}{900}\)
\(\left\{{}\begin{matrix}\dfrac{1}{48}< \dfrac{3}{32}\\\dfrac{1}{36}< \dfrac{1}{32}\\........\\\dfrac{1}{900}< \dfrac{1}{32}\end{matrix}\right.\)
Nên \(A< \dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{32}.8=1\)
chỗ kia là 1/32 mk gõ nhầm -_-