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(1-1/2)x(1-1/3)x(1-1/4)x.....x(1-1/2015)x(1-1/2016)
=> 1/2 x 2/3 x 3/4 x 4/5 x 5/6 x 6/7 x 7/8 x 8/9 x......... x 2014/2015 x 2015/2016
Ta rút gọn cho tử này mẫu kia còn: 1/2016.
Đáp số : 1/2016
\(=\frac{2-1}{2}\times\frac{3-1}{3}\times\frac{4-1}{4}\times...\times\frac{2015-1}{2015}\times\frac{2016-1}{2016}\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2014}{2015}\times\frac{2015}{2016}\)
\(=\frac{1}{2016}\)
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a)\(3\frac{2}{5}.2\frac{1}{7}=\frac{17}{5}.\frac{15}{7}=\frac{51}{7}\)
b) \(8\frac{1}{6}:2\frac{1}{2}=\frac{49}{6}:\frac{5}{2}=\frac{49}{6}.\frac{2}{5}=\frac{49}{15}\)
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a) \(0,75\times26+49\times\frac{3}{4}+\frac{3}{4}\times24+\) \(0,75\)
\(=\) \(0,75\times\left(26+1\right)+\frac{3}{4}\times\left(49+24\right)\)
\(=\) \(\frac{3}{4}\times27+\frac{3}{4}\times73\)
\(=\) \(\frac{3}{4}\times\left(27+73\right)\)
\(=\)\(\frac{3}{4}\times100=75\)
b) câu b,c làm tương tự
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Ta có:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
Nên phương trình ban đầu tương đương với:
\(\frac{5}{6}=\frac{x}{6}\Leftrightarrow x=5\)
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\(\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}\)
\(=\frac{1}{2\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
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1) =1/2 x 2/3 x 3/4 x 4/5 x .... x 2002/2003 x 2003/2004
=1/2004
2) 1/2 x X-3/4=5/6
1/2 x X =3/4+5/6
1/2 x X =19/12
X=19/6
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}=\frac{1}{2004}\)
\(\frac{1}{2}.x-\frac{3}{4}=\frac{5}{6}\)
\(\frac{1}{2}.x=\frac{5}{6}+\frac{3}{4}\)
\(\frac{1}{2}.x=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)
\(x=\frac{19}{12}:\frac{1}{2}\)
\(x=\frac{19}{12}.2=\frac{19}{6}\)
\(A=\frac{1}{2}.\frac{2}{3}\frac{3}{4}.\frac{4}{5}..\frac{5}{6}\)
\(=\frac{1}{6}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{2}{3}\right)\left(1-\frac{3}{4}\right)\left(1-\frac{4}{5}\right)\left(1-\frac{5}{6}\right)\)
\(=\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}.\frac{5-1}{5}.\frac{6-1}{6}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}=\frac{1.2.3.4.5}{2.3.4.5.6}=\frac{1}{6}\)