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TA CÓ:\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}\)
\(=1-\frac{1}{8}=\frac{7}{8}\)
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A = \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
A = \(\frac{1}{2}+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)
A = \(\left(1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
A = \(9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
A = \(9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
A = \(9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}\)
A = \(\frac{81}{10}\)
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)=10-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)=10-\left(1-\frac{1}{10}\right)=10-\frac{9}{10}=\frac{91}{10}\)
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Có công thức \(\dfrac{x}{a\left(a+x\right)}=\dfrac{1}{a}-\dfrac{1}{a+x}\) nhé!
Ví dụ: \(\dfrac{2}{2.4}=\dfrac{1}{2}-\dfrac{1}{4}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=1-\dfrac{1}{8}=\dfrac{7}{8}\)
Dấu . tức là nhân nhé!
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A = \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+......+\frac{109}{110}\)
A = \(1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}....+1-\frac{1}{110}\)
A = \(10-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)
A = \(10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)
A = \(10-\left(1-\frac{1}{11}\right)\)
A = \(10-\frac{10}{11}\)
A = \(\frac{100}{11}\)
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D=\(\frac{1}{10}+\frac{4}{20}+\frac{9}{30}+\frac{25}{50}+\frac{36}{60}+\frac{49}{70}+\frac{64}{80}+\frac{81}{90}\)
=> D=\(\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}\)
=> D = \(\frac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)
=> D =\(\frac{45}{10}=4,5\)
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Thời gian làm xong 1 dụng cụ của một học sinh học nghề là :
3 giờ 15 phút = 195 phút
195 : 5 = 39 (phút)
Thời gian làm xong 1 dụng cụ của một công nhân là :
195 : 6 = 32.5 (phút)
Vậy thời gian khi làm một dụng cụ thì công nhân làm ít hơn học sinh là :
39 - 32.5 = 6.5 (phút)
Đáp số : 6.5 phút
JshhrurieijdbfbfnzkKavskbzbb. Ơlwjwnxlw an toàn quan đoàn lạp ứa tỷ ưa liên tiếp siêu tốc khá sức ép 9
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M = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
M = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
M = 1 -\(\frac{1}{9}\)=\(\frac{8}{9}\)
A=1/2+1/6+....+1/56+1/72
A=1/1.2+1/2.3+...+1/7.8+1/8.9
A=1/1-1/2+1/2-1/3+...+1/7-1/8+1/8-1/9
A=1/1-1/9=9/9-1/9=8/9
Thanks ban nha