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A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)
B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
1)Tính
a)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
b)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
2) tìm x
\(a\)) \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}\)\(=\dfrac{9}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}\)
\(\dfrac{4}{5}x=0\)
\(x=0:\dfrac{4}{5}\)
\(x=0\)
b)\(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{4}\)
1. Tính:
a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{1}-\dfrac{1}{10}\)
= \(\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
b. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
= \(\dfrac{1}{1}-\dfrac{1}{100}\)
= \(\dfrac{100}{100}-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
2. Tìm x, biết:
a. \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}+\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{14}{5}\)
\(x=\dfrac{14}{5}:\dfrac{4}{5}\)
\(x=\dfrac{14}{5}.\dfrac{5}{4}\)
\(x=14.\dfrac{1}{4}\)
\(x=\dfrac{14}{4}\)
Vậy \(x=\dfrac{14}{4}\)
b. \(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{32}{20}+\dfrac{30}{20}\)
\(\dfrac{2}{5}x=\dfrac{62}{20}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{10}.\dfrac{5}{2}\)
\(x=\dfrac{31}{2}.\dfrac{2}{2}\)
\(x=\dfrac{31}{2}.1\)
\(x=\dfrac{31}{2}\)
Vậy \(x=\dfrac{31}{2}\)
bài này mk tự làm ko sao chép trên mạng
nếu thấy đúng thì tick đúng cho mk nha
a) A = \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)
A = \(\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}.\dfrac{4.4}{4.5}\)
A = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)= \(\dfrac{1}{5}\)
b) B = \(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)
B = \(\dfrac{2.3.4.5}{1.2.3.4}.\dfrac{2.3.4.5}{3.4.5.6}\)= \(\dfrac{5}{3}\)
\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)......\left(1+\dfrac{1}{2009.2011}\right)\)
\(=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}........\dfrac{4040100}{2009.2011}\)
\(=\dfrac{2.2.3.3.4.4.....................2010.2010}{1.3.2.4.3.5.................2008.2010.2009.2011}\)
\(=\dfrac{2.2010}{2011}\)
không biết đúng k nhưng cứ lm thôi
\(A=\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
\(A=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\right)\)
\(A=\left(1-\dfrac{1}{100}\right)-\left(\dfrac{\dfrac{1}{1.2}-\dfrac{1}{100.101}}{2}\right)\)
Bấm máy nha
\(B=\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+\dfrac{5}{3.4.5.6}+...+\dfrac{5}{98.99.100.101}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+...+\dfrac{101-98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4}{1.2.3.4}-\dfrac{1}{1.2.3.4}+\dfrac{5}{2.3.4.5}-\dfrac{2}{2.3.4.5}+...+\dfrac{101}{98.99.100.101}-\dfrac{98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{99.100.101}\right)\)
\(B=\dfrac{5}{3}.\dfrac{166649}{999900}\approx0,3\)
a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)
b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)