Các câu hỏi tương tự
Cho M = \(\frac{\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right)}{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)}\);
N = \(\frac{\left(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\right)}{\left(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}\right)}\)
Tìm tỉ số phần trăm của M và N
Bài 1: Tính nhanh
1) \(\left(1-\frac{1}{5}\right)\left(1-\frac{2}{5}\right)...\left(1-\frac{9}{5}\right)\)
2) \(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
3)\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{1999}}{\frac{1}{1.1999}+\frac{1}{3.1997}+...+\frac{1}{1997.3}+\frac{1}{1999.1}}\)
Tớ có cái này đố các cậua)frac{53}{101}.frac{-13}{97}+frac{53}{101}.frac{-84}{97}b)left(frac{1}{57}-frac{1}{5757}right)left(frac{1}{2}-frac{1}{3}-frac{1}{6}right)c)frac{-3^3}{25}.frac{75}{-21}.frac{50}{35}d)frac{25.48-25.18}{20.5^3}e)left(1+frac{1}{2}right)left(1+frac{1}{3}right)+left(1+frac{1}{4}right)+...+left(1+frac{1}{2003}right)f)left(1-frac{1}{2}right)left(1-frac{1}{3}right)+left(1-frac{1}{4}right)+...+left(1-frac{1}{200}right)Chúc các cậu may mắn!!
Đọc tiếp
Tớ có cái này đố các cậu
a)\(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)
b)\(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
c)\(\frac{-3^3}{25}.\frac{75}{-21}.\frac{50}{35}\)
d)\(\frac{25.48-25.18}{20.5^3}\)
e)\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2003}\right)\)
f)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{200}\right)\)
Chúc các cậu may mắn!!
16 tháng 2 2020 lúc 10:02
Tính : \(K=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right).3^5+\left(\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}.3^9\right)+...+\left(\frac{1}{3^{97}}+\frac{1}{3^{98}}+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right).3^{101}\)
\(F=1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}\left(1+2+3\right)-...-\frac{1}{101}\left(1+2+3+...+101\right)\)
Afrac{1}{5cdot8}+frac{1}{8cdot11}+....+frac{1}{xcdotleft(x+3right)}3A3cdotleft(frac{1}{5}-frac{1}{8}right)+left(frac{1}{8}-frac{1}{11}right)+.....left(frac{1}{x}-frac{1}{x+3}right)3A3cdotleft(frac{1}{5}-frac{1}{x+3}right)frac{101}{1540}3Aleft(frac{1}{5}-frac{1}{x+3}right)frac{101}{4620}frac{1}{x+3}frac{1}{5}-frac{101}{4620}frac{1}{x+3}?
Đọc tiếp
A=\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+....+\frac{1}{x\cdot\left(x+3\right)}\)
3A=\(3\cdot\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+.....\left(\frac{1}{x}-\frac{1}{x+3}\right)\)
3A=\(3\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
3A=\(\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{4620}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{101}{4620}\)
\(\frac{1}{x+3}=?\)
câu 1: tính giá trị biểu thức A=\(\frac{1}{7}\left(\frac{555}{222}+\frac{4444}{12221}+\frac{33333}{244442}+\frac{11}{330}+\frac{13}{60}\right)\)
câu 2 :Cho E=92-\(\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\) và F =\(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}\) .Tính\(\frac{E}{F}\)
Afrac{1+frac{1}{3}+frac{1}{5}+frac{1}{7}+......+frac{1}{999}}{frac{1}{1.999}+frac{1}{3.997}+frac{1}{5.995}+......+frac{1}{999.1}}Bfrac{1+left(1+2right)+left(1+2+3right)+left(1+2+3+4right)+......+left(1+2+3+...+98right)}{1.2+2.3+3.4+4.5+......+98.99}Cfrac{frac{1}{1.300}+frac{1}{2.301}+frac{1}{3.302}+......+frac{1}{100.400}}{frac{1}{1.102}+frac{1}{2.103}+frac{1}{3.104}+......+frac{1}{299.400}}Dfrac{frac{1}{99}+frac{2}{98}+frac{3}{97}+......+frac{99}{1}}{frac{1}{2}+frac{1}{3}+frac{1}{4}+......+frac...
Đọc tiếp
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
B1:Tính hợp lí
a) \(1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}\left(1+2+3\right)-...-\frac{1}{101}\left(1+2+...+101\right)\)
B2
Chứng minh \(1.3.5....99=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}\)
Giải nhanh nhé .Mình đag cần gấp .Cảm ơn!