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\(D=\dfrac{4}{8\cdot13}+\dfrac{4}{13\cdot18}+\dfrac{4}{18\cdot23}+...+\dfrac{4}{253\cdot258}\\ =\dfrac{4}{5}\cdot\dfrac{5}{8\cdot13}+\dfrac{4}{5}\cdot\dfrac{5}{13\cdot18}+\dfrac{4}{5}\cdot\dfrac{5}{18\cdot23}+...+\dfrac{4}{5}\cdot\dfrac{5}{253\cdot258}\\ =\dfrac{4}{5}\left(\dfrac{5}{8\cdot13}+\dfrac{5}{13\cdot18}+\dfrac{5}{18\cdot23}+...+\dfrac{5}{253\cdot258}\right)\\ =\dfrac{4}{5}\cdot\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\\ =\dfrac{4}{5}\cdot\left(\dfrac{1}{8}-\dfrac{1}{258}\right)\\ =\dfrac{4}{5}\cdot\dfrac{125}{1032}\\ =\dfrac{25}{258}\)
ta có
Tính:
\(\dfrac{4}{8.13}+\dfrac{4}{13.18}+....+\dfrac{4}{253.258}\)
= 4 (\(\dfrac{1}{8.13}+\dfrac{1}{13.18}+.....+\dfrac{1}{253.258}\))
=\(\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)
=\(\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{258}\right)\)
=\(\dfrac{25}{258}\)
\(B=\dfrac{1}{6.10}+\dfrac{1}{10.14}+...+\dfrac{1}{402.406}\\ 4B=\dfrac{4}{6.10}+\dfrac{4}{10.14}+...+\dfrac{4}{402.406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{402}-\dfrac{1}{406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}\\B=\dfrac{\dfrac{100}{609}}{4}=\dfrac{25}{609} \)
\(B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{ 1}{402}-\dfrac{1}{406}\)
\(=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}.\)
B=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
B=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
B=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
B= 1-\(\dfrac{1}{8}\)
B= \(\dfrac{7}{8}\)
\(A=\dfrac{5}{9}-\dfrac{5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\ =\dfrac{5}{9}+\dfrac{-5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\= \left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{-5}{8}+\dfrac{-3}{8}\right)\\ =1+1+\left(-1\right)\\ =2+\left(-1\right)\\ =1\)
a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)
b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)
c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)
d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)
lưu ý mk ko chép đầu bài
mình cần gấp lắm đến chiều mai là phải nộp rùi
giúp mình nha thanks cá bạn trước 
ko có tâm trạng mà cười nữa
\(A=\dfrac{1}{3}\cdot\dfrac{4}{5}+\dfrac{1}{3}\cdot\dfrac{6}{5}+\dfrac{2}{3}\\ =\dfrac{1}{3}\cdot\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\\ =\dfrac{1\cdot2}{3}+\dfrac{2}{3}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{4}{3}\)
B =\(\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}\)-\(\dfrac{40}{57}\)
=\(\dfrac{4}{19}.\left[\dfrac{-5}{6}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)
=\(\dfrac{4}{19}.\left[\dfrac{-10}{12}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)
=\(\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)
=\(\dfrac{-17}{57}\)-\(\dfrac{40}{57}\)
=\(\dfrac{-17}{57}+\dfrac{-40}{57}\)
=\(\dfrac{-57}{57}=-1\)
a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
a, \(\dfrac{-7}{9}.2\dfrac{3}{4}\)
= \(\dfrac{-7}{9}.\dfrac{11}{4}\)
= \(\dfrac{-77}{36}\)
b, \(\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{-2}{5}\)
= \(\dfrac{2}{3}+\dfrac{-2}{15}\)
= \(\dfrac{10}{15}+\dfrac{-2}{15}\)
= \(\dfrac{-8}{15}\)
c , \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\)
= \(\dfrac{2}{3}-4.\dfrac{5}{4}\)
= \(\dfrac{2}{3}-5\)
= \(\dfrac{-13}{3}\)
d, \(\left(\dfrac{1}{-3}+\dfrac{5}{6}\right).11-7\)
= \(\dfrac{1}{2}\) . 11 - 7
= \(\dfrac{11}{2}-\dfrac{14}{2}\)
= \(\dfrac{-3}{2}\)
e, \(\dfrac{3}{4}.15\dfrac{1}{3}-\dfrac{3}{4}.43\dfrac{1}{3}\)
= \(\dfrac{3}{4}.\left(15\dfrac{1}{3}-43\dfrac{1}{3}\right)\)
= \(\dfrac{3}{4}.-28\)
= \(-21\)
Cho A = 1/2 .3/4.5/6.....199/200.Chứng tỏ rằng B mũ 2 <1/201.Bạn có làm dược ko ?
a , 2/8 - 2/13 + 2/13 - 2/18 + 2/18 - 2/23 + ... + 2/253 - 2/258
= 2/8 - 2/258
= 125/516
b , 1/6 - 1/10 + 1/10 - 1/14 + 1/14 - 1/18 + ... + 1/402 - 1/406
= 1/6 - 1/406
= 100/609