A = ( \(\dfrac{1}{2}\) + 1)(\(\dfrac{1}{3}\) + 1).(\(\dfrac{1}{4}\) + 1)....(\(\dfrac{1}{999}\) + 1)
A = \(\dfrac{3}{2}\).\(\dfrac{4}{3}\).\(\dfrac{5}{4}\).......\(\dfrac{1000}{999}\)
A = \(\dfrac{3.4.5......999}{3.4.5......999}\). \(\dfrac{1000}{2}\)
A = 1 \(\times\) 500
A = 500
B=(1-\(\dfrac{1}{2^2}\))(1-\(\dfrac{1}{3^2}\))(1-\(\dfrac{1}{4^2}\))(1-\(\dfrac{1}{5^2}\))...(1-\(\dfrac{1}{99^2}\))(1-\(\dfrac{1}{100^2}\))
C=(\(\dfrac{1}{4}\)-1)(\(\dfrac{1}{9}\)-1)(\(\dfrac{1}{16}\)-1)...(\(\dfrac{1}{999}\)-1)
1.Tính nhanh
A =( \(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\)) x \(\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
2.Tìm n ∈ Z để \(\dfrac{n+3}{n-2}\)nhận giá trị nguyên
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
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) \(5\dfrac{1}{2}-\left(\dfrac{1}{4}.x+\dfrac{2}{5}\right)=25\%\)
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 , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
Chứng tỏ rằng:
a)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
b)\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
\(\dfrac{x-1}{21}=\dfrac{3}{x+1}\)
\(2\dfrac{1}{2}x+x=2\dfrac{1}{17}\)
\(\left(x+\dfrac{1}{4}-\dfrac{2}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(2\dfrac{1}{3}x-1\dfrac{3}{4}x+2\dfrac{2}{3}=3\dfrac{3}{5}\)
Giúp mình với ! Mình cần gấp
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
94-\(\dfrac{1}{7}-\dfrac{2}{8}-\dfrac{3}{9}-...-\dfrac{94}{100}\)
\(\dfrac{1}{35}+\dfrac{1}{40}+\dfrac{1}{45}+...+\dfrac{1}{500}\)
giúp mik nha mik cần gấp
