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A=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
=\(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{49}-\frac{2}{51}\)
= \(2.(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51})\)
=2.\((1-\frac{1}{51})\)
=\(2.\frac{50}{51}\)
=\(\frac{100}{51}\)
......................?
mik ko biết
mong bn thông cảm
nha ................
−12−−35+−19+1131−−27+435−718−12−−35+−19+1131−−27+435−718
=−12+35−19+1131+27+435−718=−12+35−19+1131+27+435−718
=110−1221179+25−718=110−1221179+25−718
=−4111790+190=−4111790+190
=1131
Bài 1:
\(=\dfrac{-1}{2}+\dfrac{3}{5}-\dfrac{1}{9}+\dfrac{1}{131}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\)
\(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{131}\)
\(=\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{131}\)
=1/131
Bài 2:
b: \(B=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)
Câu a) bạn tham khảo tại đây nhé: Câu hỏi của Hằng Thanh.
Chúc bạn học tốt!
\(\frac{-1}{2}-\frac{-3}{5}-\frac{1}{9}+\frac{1}{131}+\frac{2}{7}+\frac{4}{35}-\frac{7}{18}=\left(-\frac{1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{2}{7}+\frac{4}{35}\right)+\frac{1}{131}\)\(=\left(\frac{-9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{10}{35}+\frac{4}{35}\right)+\frac{1}{131}=\frac{-18}{18}+\frac{35}{35}+\frac{1}{131}=-1+1+\frac{1}{131}=\frac{1}{131}\)