Tính:
a) \(A=-3+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}\)
Tính nhanh
a, S= \(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\) + \(\dfrac{1}{99}\) + \(\dfrac{1}{143}\)
b, A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\)
c, H =\(\dfrac{4047991-2010x2009}{4050000-2011x2009}\)
d, T = \(\dfrac{2009x20010+2000}{2011x2010-2020}\)
e, P = \(\dfrac{7589-80,5x69,3}{7485,05-79x69,3}\)
f, B = 5,1 x 42,2 + 1,7 x 448 x 3 - 0,15 x 700
Giúp mình với
a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
1. Thực hiện phép tính A=3.\(\dfrac{1}{1\cdot2}\)- 5.\(\dfrac{1}{2\cdot3}\)+7.\(\dfrac{1}{3\cdot4}\)- ... +15\(\dfrac{1}{7\cdot8}\)-17\(\dfrac{1}{8\cdot9}\)
2.Tính tỉ số \(\dfrac{A}{B}\) biết A=\(\dfrac{1}{1\cdot300}\)+\(\dfrac{1}{2\cdot301}\)+\(\dfrac{1}{3\cdot302}\)+...+\(\dfrac{1}{101\cdot400}\) và B=\(\dfrac{1}{1\cdot102}\)+\(\dfrac{1}{2\cdot103}\)+\(\dfrac{1}{3\cdot104}\)+...+\(\dfrac{1}{299\cdot400}\)
Nhanh lên nhé, vội lắm rồi
Tính : \(A=-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{101}}\)
`3A=-1+1/3-1/3^2+.....+1/3^99-1/3^100`
`=>3A+A=4A=-1-1/3^101`
`=>A=(-1-1/3^101)/4`
Bài 1: tính
a) 3\(\dfrac{1}{2}\) + 4\(\dfrac{5}{7}\) - 5\(\dfrac{5}{14}\) b) 4\(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) : \(5\dfrac{1}{2}\)
bài 2: tìm X
a) X x \(3\dfrac{1}{3}\) = \(3\dfrac{1}{3}\) : \(4\dfrac{1}{4}\) b) \(5\dfrac{2}{3}\) : X = \(3\dfrac{2}{3}\) - \(2\dfrac{1}{2}\)
các giáo viên olm giúp e vs, e cần gấp lắm!
\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)
= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)
= \(\dfrac{40}{14}=\dfrac{20}{7}\)
\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)
=\(\dfrac{9}{2}+\dfrac{1}{11}\)
=\(\dfrac{101}{22}\)
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)
\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)
\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)
\(x=\dfrac{102}{21}=\dfrac{34}{7}\)
cho A=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2022}\)
B=\(\dfrac{2021}{1}+\dfrac{2020}{2}+\dfrac{2019}{3}+...+\dfrac{1}{2021}\)
tính tỉ số \(\dfrac{B}{A}\)
\(B=\left(\dfrac{2020}{2}+1\right)+\left(\dfrac{2019}{3}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1\)
\(=\dfrac{2022}{2}+\dfrac{2022}{3}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}\)
=2022(1/2+1/3+...+1/2021+1/2022)
=>B/A=2022
Cho \(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2022}\)
Và \(B = \dfrac{2021}{1}+\dfrac{2020}{2}+\dfrac{2019}{3}+...+\dfrac{1}{2021}\)
Tính B/A
Tính A biết
A=1- \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)-\(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{102}}\)-\(\dfrac{1}{3^{103}}\)
\(3A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{103}}-\dfrac{1}{3^{104}}\)
=>\(4A=3-\dfrac{1}{3^{104}}=\dfrac{3^{105}-1}{3^{104}}\)
=>\(A=\dfrac{3^{105}-1}{3^{104}\cdot4}\)
\(3A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{103}}-\dfrac{1}{3^{104}}\)
=>\(4A=3-\dfrac{1}{3^{104}}=\dfrac{3^{105}-1}{3^{104}}\)
=>\(A=\dfrac{3^{105}-1}{3^{104}\cdot4}\)
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
Tính hợp lí :
A = \(\dfrac{-2}{9}\) + \(\dfrac{-3}{4}\) + \(\dfrac{3}{5}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{57}\) + \(\dfrac{1}{3}\) + \(\dfrac{-1}{36}\)
B = \(\dfrac{1}{2}\) + \(\dfrac{-1}{5}\) + \(\dfrac{-5}{7}\) + \(\dfrac{1}{6}\) + \(\dfrac{-3}{35}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{41}\)
C = \(\dfrac{-1}{2}\) + \(\dfrac{3}{5}\) + \(\dfrac{-1}{9}\) + \(\dfrac{1}{127}\) + \(\dfrac{-7}{18}\) + \(\dfrac{4}{35}\) + \(\dfrac{2}{7}\)
Cho:
\(A=\dfrac{1}{1.\left(2n-1\right)}+\dfrac{1}{3.\left(2n-3\right)}+...+\dfrac{1}{\left(2n-3\right).3}+\dfrac{1}{\left(2n-1\right).1}\) \(B=1+\dfrac{1}{3}+...+\dfrac{1}{2n-1}\) (với n ∈ N*).
Tính \(\dfrac{A}{B}\)