tính GTBT:
a)a.\(\left(\dfrac{-3}{2}\right)+a.\dfrac{1}{4}-a.\dfrac{5}{6}vớia=\dfrac{3}{5}\)
b)\(\dfrac{2}{5}.b-\dfrac{1}{3}.b+b.\left(\dfrac{-1}{2}\right)vớib=\dfrac{6}{13}\)
c)c.\(\dfrac{5}{6}+\dfrac{3}{4}.c-\dfrac{11}{12}.c\) với c=\(\dfrac{2019}{2020}\)
Chứng minh rằng: \(\dfrac{1}{201}+\dfrac{1}{202}+\dfrac{1}{203}+\dots+\dfrac{1}{400}< \dfrac{5}{6}\)
Bài 7:
Chứng minh rằng: \(\dfrac{3}{10}< \dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}.\)
Giúp em với ạ!!!
Thực hiện các phép tính:
a/ \(\dfrac{2}{3}\)+\(\dfrac{1}{5}\).\(\dfrac{10}{7}\)\(\)
b/ \(\dfrac{7}{12}\)-\(\dfrac{27}{7}\)\(\).\(\dfrac{1}{8}\)
c/ \(\dfrac{5}{9}\).\(\dfrac{7}{13}\)\(\)+\(\dfrac{5}{9}\).\(\dfrac{9}{13}\)-\(\dfrac{3}{13}\).\(\dfrac{5}{9}\)
Em cần ngay ạ!!!
chứng minh rằng : \(\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)
Chứng minh rằng:
\(\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{99^2}< \dfrac{5}{18}\)
\(\dfrac{3}{2}X-0,2=\dfrac{3}{5}\)
\(\dfrac{1}{3}+x=\dfrac{3}{4}\)
\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{11}{8}-\dfrac{3}{8}.x=\dfrac{1}{8}\)
giúp với
Tìm x
a/\(\dfrac{x}{5}\)+\(\dfrac{1}{2}\)=\(\dfrac{6}{10}\)
b/\(\dfrac{1}{2}\).\(x\)+\(\dfrac{1}{2}\)=\(\dfrac{5}{2}\)
c/\(\dfrac{1}{2}\)-\(\dfrac{2}{3}\).\(x\)=\(\dfrac{7}{12}\)
giúp e ạ
1, \(\dfrac{3}{4}\). ( \(\dfrac{2}{5}\) - \(\dfrac{1}{15}\)) +\(\dfrac{3}{4}\)
2, \(\dfrac{4}{9}\). (\(\dfrac{-13}{3}\)) + \(\dfrac{4}{3}\). \(\dfrac{40}{9}\)
3, \(\dfrac{4}{9}\) - \(\dfrac{2}{3}\). ( \(\dfrac{4}{5}\)+\(\dfrac{1}{2}\) )
giúp mình nha cảm ơn