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31 tháng 7 2015

 Trương Quang Minh

31 tháng 7 2015

                        

4 tháng 8 2018

mk chỉnh lại đề

\(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}....\frac{99}{100}\)

\(=\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.\frac{5^2-1}{5^2}....\frac{10^2-1}{10^2}\)

\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{9.11}{10^2}\)

\(=\frac{2.3.4...9}{3.4.5...10}.\frac{4.5.6...11}{3.4.5...10}\)

\(=\frac{2}{10}.\frac{11}{3}=\frac{11}{15}\)

21 tháng 7 2016

(1 - 1/3) × (1 - 1/6) × (1 - 1/10) × (1 - 1/15) × ... × (1/780) × a = 1

2/3 × 5/6 × 9/10 × 14/15 × ... × 779/780 × a = 1

4/6 × 10/12 × 18/20 × 28/30 × ... × 1558/1560 × a = 1

1×4/2×3 × 2×5/3×4 × 3×6/4×5 × 4×7/5×6 × ... × 38×41/39×40 × a = 1

1×2×3×4×...×38/2×3×4×5×...×39 × 4×5×6×7×...×41/3×4×5×6×...×40 × a = 1

1/39 × 41/3 × a = 1

41/297 × a = 1

=> a = 297/41

21 tháng 7 2016

1 - 1/3) × (1 - 1/6) × (1 - 1/10) × (1 - 1/15) × ... × (1/780) × a = 1

2/3 × 5/6 × 9/10 × 14/15 × ... × 779/780 × a = 1

4/6 × 10/12 × 18/20 × 28/30 × ... × 1558/1560 × a = 1

1×4/2×3 × 2×5/3×4 × 3×6/4×5 × 4×7/5×6 × ... × 38×41/39×40 × a = 1

1×2×3×4×...×38/2×3×4×5×...×39 × 4×5×6×7×...×41/3×4×5×6×...×40 × a = 1

1/39 × 41/3 × a = 1

41/297 × a = 1

=> a = 297/41

11 tháng 8 2016

\(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right).a=1\)

=> \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}.a=1\)

=> \(\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}.a=1\)

=> \(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{38.41}{39.40}.a=1\)

=> \(\frac{1.2.3.4...38}{2.3.4.5...39}.\frac{4.5.6.7...41}{3.4.5.6...40}.a=1\)

=> \(\frac{1}{39}.\frac{41}{3}.a=1\)

=> \(\frac{41}{117}.a=1\)

=> \(a=1:\frac{41}{117}=\frac{117}{41}\)

31 tháng 5 2017

117/41

15 tháng 12 2015

ai cho mình thêm 4 li-ke cho lên 155 với

20 tháng 8 2017

\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)

\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)

\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)

\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)

làm giống như trên

\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)

\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)

P/S: . là nhân nha

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