ta thấy : \(\dfrac{-1003}{-2002}\) = \(\dfrac{1003}{2002}\)
\(\dfrac{1004}{-2003}\) = \(\dfrac{-1004}{2003}\)
Sắp xếp : \(\dfrac{1004}{-2003}\) <\(\dfrac{-1003}{2003}\) <\(\dfrac{-1002}{2003}\) <\(\dfrac{1001}{2002}\) <\(\dfrac{-1003}{-2002}\)
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
ta thấy : \(\dfrac{-1003}{-2002}\) = \(\dfrac{1003}{2002}\)
\(\dfrac{1004}{-2003}\) = \(\dfrac{-1004}{2003}\)
Sắp xếp : \(\dfrac{1004}{-2003}\) <\(\dfrac{-1003}{2003}\) <\(\dfrac{-1002}{2003}\) <\(\dfrac{1001}{2002}\) <\(\dfrac{-1003}{-2002}\)
Các phân số sau đây được sắp xếp theo một quy luật. Hãy quy đồng mẫu các phân số để tìm quy luật đó rồi điền tiếp vào chỗ trống một phân số thích hợp :
a) \(\dfrac{1}{6};\dfrac{1}{3};\dfrac{1}{2};.....\)
b) \(\dfrac{1}{8};\dfrac{5}{24};\dfrac{7}{24};....\)
c) \(\dfrac{1}{5};\dfrac{1}{4};\dfrac{3}{1};...\)
d) \(\dfrac{4}{15};\dfrac{3}{10};\dfrac{1}{3};....\)
câu 4 : chứng minh rằng:
1 phần 2 mũ 2 + 1 phần 3 mũ 2 + .................... + 1 phần 2002 mũ 2 + 1 phần 2003 mũ 2 <1
tính H = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}:\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\)
\(\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)
Bài 1. Thực hiện phép tính:
a) |5.0,6+\(\dfrac{2}{3}\)|- \(\dfrac{1}{3}\)
b)(0,25 - 1\(\dfrac{1}{4}\)) : 5 - \(\dfrac{1}{5}\).(-3)\(^2\)
c)\(\dfrac{14}{17}.\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
d)\(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
e)\(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
Chứng tỏ rằng:
a) \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{2}\)
b) \(S=\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{80}>\dfrac{7}{12}\)
c) \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
d) \(\dfrac{49}{100}< S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}< 1\)
Các bạn giải ra từng bước dùm mik nha
Thanks m.n
chứng minh :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}\) b) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
So sánh \(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\) và \(B=2018\)