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1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000
= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000
= 1/10 - 1/1000
= 100/1000 - 1/1000
= 99/1000
1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000
= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000
= 1/10 - 1/1000
= 100/1000 - 1/1000
= 99/1000
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3C = 1x2x3 + 2x3x(4-1) + ... + 11x12x(13-9)
3C = 1x2x3 + 2x3x4 - 1x2x3 + ... + 11x12x13 - 9x11x12
3C = 9x11x12
3C = 1188
C = 396
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\(\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+....+\frac{5}{49.50}\)
\(=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+.....+\frac{1}{49.50}\right)\)
\(=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(=5.\left(\frac{1}{10}-\frac{1}{50}\right)\)
\(=5.\frac{2}{25}\)
\(=\frac{2}{5}\)
5/10 x 11 + 5/11 x 12 + 5/12 x 13 + .... + 5/49 x 50
= 5/10 - 5/11 + 5/11 - 5/12 + 5/12 - 5/13 + ....... + 5/49 - 5/50
= 5/10 - 5/50
= 2/5
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Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
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\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
= 7/10-7/11+7/11-7/12+7/12-7/13+...+7/69-7/70
=7/10-7/70
=42/70
k mk nha
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Bài nào khó lắm thì mới hỏi thôi chứ bài này dễ mà bạn tự vận động não đi
\(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{11\cdot12}=x\)
\(\Leftrightarrow x=\dfrac{1}{3}-\dfrac{1}{12}=\dfrac{4}{12}-\dfrac{1}{12}=\dfrac{1}{4}\)
9/100