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\(E=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-1\frac{2}{7}\right).\left(1-\frac{3}{7}\right)\)
\(E=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)...\left(1-1\frac{2}{7}\right).\left(1-\frac{3}{7}\right)\)
\(E=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...0...\left(1-1\frac{2}{7}\right).\left(1-\frac{3}{7}\right)\)
\(E=0\)
\(E=\frac{7-1}{7}+\frac{7-2}{7}+\frac{7-3}{7}+...+\frac{7-9}{7}+\frac{7-10}{7}\)
Vì trong biểu thức E có số hạng \(\frac{7-7}{7}=0\)
Nên E=0 (ĐPCM)
hok tốt
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
\(\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\left(1-1\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{3}{7}\right).0\)
\(=0\)
Trong dãy nhất định có \(\left[1-\frac{7}{7}\right]=0\)nên tích dãy trên là 0
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2}{3}...\frac{2017}{2018}\)
\(=\frac{1.2...2017}{2.3...2018}\)
\(=\frac{1}{2018}\)
b) \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{190}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{189}{190}\)
\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{378}{380}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{7.4}{5.6}...\frac{18.21}{19.20}\)
\(=\frac{\left(1.2.3...18\right).\left(4.5.6...21\right)}{\left(2.3.4...19\right).\left(3.4.5...20\right)}\)
\(=\frac{1.21}{19.3}\)
\(=\frac{21}{57}\)
c) \(\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)\left(1+\frac{7}{48}\right)...\left(1+\frac{7}{2009}\right)\)
\(=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}.\frac{56}{48}...\frac{2016}{2009}\)
mk ko bít làm câu c ! xin lỗi bn nha! bn tự nghĩ cách làm câu c giúp mk nhé!
\(E=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-1\frac{2}{7}\right).\left(1-\frac{3}{7}\right)\)
\(E=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)...\left(1-1\frac{2}{7}\right).\left(1-\frac{3}{7}\right)\)
\(E=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...0...\left(1-1\frac{2}{7}\right).\left(1-\frac{3}{7}\right)\)
\(E=0\)