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phần a dễ bạn tự làm đi tử thì bạn tính như bình thường còn mẫu thì:7.(\(\frac{1}{3.13}\)+\(\frac{1}{13.23}\)+\(\frac{1}{23.33}\))
\(\frac{7}{10}\).(\(\frac{1}{3}\)-\(\frac{1}{33}\))=\(\frac{7}{33}\)
b)(1+1/3+1/5+..+1/199)-(1/2+1/4+...+1/200)
(1+1/2+1/3+...+1/199+1/200)-(1/2+1/2+1/4+1/4+...+1/200+1/200)
=1+1/2+1/3+...+1/199+1/200-(1+1/2+1/3+...+1/100)
=1/101+1/102+...+1/200
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!
Ta có :
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}\)
\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1.2.2.3.3.4.....99.100}\)
\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1^2.2^2.3^2.4^2.....99^2}.\frac{1}{100}\)
\(=\)\(\frac{1}{100}\)
a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)
b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0
a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}\)
\(=\frac{25}{33}\)
b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)
Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.
\(\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\left(1-1\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{3}{7}\right).0\)
\(=0\)
Trong dãy nhất định có \(\left[1-\frac{7}{7}\right]=0\)nên tích dãy trên là 0
Bài 1:
\(\Leftrightarrow-\dfrac{5}{7}:x=-\dfrac{7}{18}-\dfrac{1}{6}=\dfrac{-7}{18}-\dfrac{3}{18}=\dfrac{-10}{18}=\dfrac{-5}{9}\)
=>x=5/9:5/7=7/9
Bài 2:
a: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{1000}{999}=\dfrac{1000}{2}=500\)
b: \(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-999}{1000}\)
\(=-\dfrac{1}{1000}\)
\(B=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)
\(B=10.\left(\frac{1}{3.13}+\frac{1}{13.23}+....+\frac{1}{53.63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}+\frac{1}{63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(B=10.\frac{20}{63}\)
\(B=\frac{200}{63}\)
Cảm ơn bn nha