11/1x4+11/4x7+11/7x10+...11/100x103
1/5x9+1/9x13+1/13x17+...+1/2450
mình cần gấp nhé
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)
\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)
\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)\)
\(=2.\dfrac{99}{100}\)
\(=\dfrac{99}{50}\)
_____
b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)
\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)
\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=3.\left(1-\dfrac{1}{101}\right)\)
\(=3.\dfrac{100}{101}\)
\(=\dfrac{300}{101}\)
\(-\frac{12}{35}\div\frac{7}{11}-\frac{23}{35}\div\frac{7}{11}-\frac{5}{11}\)
\(=\left(-\frac{12}{35}-\frac{23}{35}\right)\div\frac{7}{11}-\frac{5}{11}\)
\(=-1\div\frac{7}{11}-\frac{5}{11}\)
\(=-\frac{11}{7}-\frac{5}{11}\)
\(=-\frac{156}{77}\)
\(S=\frac{1}{4}\times\left(\frac{4}{5\times9}+\frac{4}{9\times13}+\frac{4}{13\times17}+...+\frac{4}{41\times45}\right)\)
\(S=\frac{1}{4}\times\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(S=\frac{1}{4}\times\left(\frac{1}{5}-\frac{1}{45}\right)\)
\(S=\frac{1}{4}\times\frac{8}{45}\)
\(S=\frac{1\times2}{1\times45}\)
\(S=\frac{2}{45}\)
Vậy \(S=\frac{2}{45}\)
Tk nha bn !!
a) 1/1x5 + ... + 1/21x25
= 4 x (1-1/5 + 1/5 - 1/9 + ... + 1/21 - 1/25)
= 1/4 x (1 - 1/25)
= 1/4 x 24/25
= 6/25
\(A=\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+...+\frac{7}{17\cdot21}=\)
\(\frac{4}{7}A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{17\cdot21}=\)
\(\frac{4}{7}A=\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+...+\left(\frac{1}{17}-\frac{1}{21}\right)=\)
\(\frac{4}{7}A=1-\frac{1}{21}=\)
\(\frac{4}{7}A=\frac{20}{21}\)
\(A=\frac{20}{21}\div\frac{4}{7}\)
\(A=\frac{20}{21}\times\frac{7}{4}=\frac{140}{84}=\frac{5}{3}\)
\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+...+\frac{7}{17.21}\)
\(=\frac{7}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{17.21}\right)\)
\(=\frac{7}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{21}\right)\)
\(=\frac{7}{4}.\left(1-\frac{1}{21}\right)\)
\(=\frac{7}{4}.\frac{20}{21}=\frac{5}{3}\)
\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)
\(=\frac{11}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{11}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=\frac{11}{3}\left(1-\frac{1}{103}\right)\)
Tự tính
\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)
= \(\frac{11}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
= \(\frac{11}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
= \(\frac{11}{3}.\left(1-\frac{1}{103}\right)\)
= \(\frac{11}{3}.\frac{102}{103}\)
= \(\frac{374}{103}\)