\(\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\\ =-\dfrac{5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\\ =-\dfrac{5}{8}+\dfrac{1}{2}=-\dfrac{5}{8}+\dfrac{4}{8}=-\dfrac{1}{8}\)
S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}...+\dfrac{1}{9.10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{2}{5}\)
S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9}< 1-\dfrac{1}{9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{8}{9}\)
Vậy \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
a) 0,25-\(\dfrac{2}{3}\)+1\(\dfrac{1}{4}\)
b) \(\dfrac{3^2}{2}\):\(\dfrac{1}{4}\)+\(\dfrac{3}{4}\).2010
c) {[(\(\dfrac{1}{25}\)-0,6)2:\(\dfrac{49}{125}\)].\(\dfrac{5}{6}\)}-[(\(\dfrac{-1}{3}\))+\(\dfrac{1}{2}\)]
d) (-\(\dfrac{1}{2}\)-\(\dfrac{1^{ }}{3}\))2:[(\(\dfrac{-5}{36}\))-(\(\dfrac{-5}{36}\))0]
Mn giúp mk nhé mk gấp quá tí đi học ai làm được mk thả tim và like nhé
\(\dfrac{x-1}{21}=\dfrac{3}{x+1}\)
\(2\dfrac{1}{2}x+x=2\dfrac{1}{17}\)
\(\left(x+\dfrac{1}{4}-\dfrac{2}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(2\dfrac{1}{3}x-1\dfrac{3}{4}x+2\dfrac{2}{3}=3\dfrac{3}{5}\)
Giúp mình với ! Mình cần gấp
Bài 1.(2,5 điểm)Tìm x, biết:
a) \(\left(2\dfrac{1}{3}+3\dfrac{1}{2}\right).x=-4\dfrac{1}{6}+3\dfrac{1}{2}\)
b) \(\left(1\dfrac{1}{3}+3\dfrac{1}{2}\right).x=4\dfrac{1}{6}-3\dfrac{1}{2}\)
c) \(\dfrac{1}{3}-\dfrac{7}{8}.x=\dfrac{1}{4}\)
d) \(\dfrac{3}{2}.x+\dfrac{1}{7}=\dfrac{7}{8}.\dfrac{64}{49}\)
e) \(5\dfrac{1}{2}-\left(\dfrac{1}{4}.x+\dfrac{2}{5}\right)=25\%\)
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
1) \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
2) \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(0,8-\dfrac{3}{4}\right)^2\)
3) \(16\dfrac{2}{7}:\left(\dfrac{-3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
4) \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
5) \(\left(\dfrac{1}{3}\right)^{50}.\left(-9\right)^{25}-\dfrac{2}{3}:4\)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
-\(\dfrac{4}{7}\)+\(\dfrac{15}{4}\)-(\(\dfrac{11}{4}\)+\(\dfrac{3}{7}+\dfrac{1}{2}-\dfrac{1}{3}\))
\(\dfrac{1}{5}\left(\dfrac{1}{2}-\dfrac{1}{3}\right):\left(\dfrac{-9}{10}\right)+\dfrac{-7}{3}\)
\(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)
giúp mik với
tìm x
\(x-\dfrac{1}{2}=\dfrac{3}{4}\)
\(x+\dfrac{7}{8}=\dfrac{3}{4}\)
\(\dfrac{1}{2}.x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\dfrac{1}{2}-\dfrac{5}{6}:x=\dfrac{2}{3}\)
Chứng tỏ rằng:
a)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
b)\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
