Đại số lớp 6
Các câu hỏi tương tự
Tìm x
(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+...+\(\dfrac{1}{2013}\)). x +2013=\(\dfrac{2014}{1}\)+\(\dfrac{2015}{2}\)+...+\(\dfrac{4024}{2012}\)+\(\dfrac{4026}{2013}\)
Tính:
a, A = 1+2-3-4+5+6-7-8 +........+2013+2014
b, B = (1+\(\dfrac{1}{2}\) ) . ( 1+\(\dfrac{1}{3}\) ) . ( 1+\(\dfrac{1}{4}\) ) ....... (1+\(\dfrac{1}{2015}\))
cho A=\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\)
So sánh A với\(\dfrac{1}{4}\)
Tính \(\dfrac{P}{A}\)biết :
P=\(\dfrac{2013}{2}+\dfrac{2013}{3}+\dfrac{2013}{4}+...+\dfrac{2013}{2014}\)
A = \(\dfrac{2013}{1}+\dfrac{2012}{2}+\dfrac{2011}{1}+....+\dfrac{1}{2013}\)
Cho A\(=\dfrac{\left(3\dfrac{2}{15}+\dfrac{1}{5}\right):2\dfrac{1}{2}}{\left(5\dfrac{3}{7}-2\dfrac{1}{4}\right):4\dfrac{43}{56}}\)
B\(=\dfrac{1,2:\left(1\dfrac{1}{5}\cdot1\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
Chứng tỏ A=B
Chứng tỏ A<1 biết :
A=\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{2013}{2014!}\)
Bài 1 : Tính
a) Adfrac{dfrac{1}{9}-dfrac{1}{7}-dfrac{1}{11}}{dfrac{4}{9}-dfrac{4}{7}-dfrac{4}{11}}+dfrac{0,6-dfrac{3}{25}-dfrac{3}{125}-dfrac{3}{625}}{dfrac{4}{5}-0,16-dfrac{4}{125}-dfrac{4}{625}}
b)Bdfrac{8}{9}-dfrac{1}{72}-dfrac{1}{56}-dfrac{1}{42}-...-dfrac{1}{6}-dfrac{1}{2}
c)Cdfrac{1}{1.3}-dfrac{1}{2.4}+dfrac{1}{3.5}-dfrac{1}{4.6}+dfrac{1}{5.7}-dfrac{1}{6.8}+dfrac{1}{7.9}-dfrac{1}{8.10}
1 tiếng nữa mình cần rồi giúp mình nhé !!!!!!!!!! Vạn lời tri ân
Đọc tiếp
Bài 1 : Tính
a) A=\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}\)\(+\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)
b)B=\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)
c)C=\(\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)
1 tiếng nữa mình cần rồi giúp mình nhé !!!!!!!!!! Vạn lời tri ân
tính giá trị biểu thức sau
A= -1 -2 +3 +4 -5 -6 +7 +8-9-10+11+12 -...........-2013-2014+2015+2016
B= [\(\dfrac{1}{2}\) -1 ] : [\(\dfrac{1}{3}\) -1] : ...................: [\(\dfrac{1}{99}\) -1} : [\(\dfrac{1}{100}\) -1]
A= 1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+........+\dfrac{1}{3^{2014}}\)
So sánh A với \(\dfrac{3}{2}\)