Các câu hỏi tương tự
Tính tổng :
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
THỰC HIỆN PHÉP TÍNH SAU
50 + \(\frac{50}{3}\)+ \(\frac{25}{3}\)+ \(\frac{20}{4}\)+\(\frac{10}{3}\)+\(\frac{100}{6.7}\)+.........+ \(\frac{100}{98.99}\)+\(\frac{1}{99}\)
( GIẢI CHI TIẾT MÌNH LIKE CHO )
Tính tổng:
\(50+\frac{50}{3}+\frac{25}{3}+\frac{30}{4}+\frac{10}{3}+\frac{100}{6.7}+...+198.99\)
Tính tổng B=50+50/3+25/3+20/4+10/3+100/6.7+...+100/98.99+1/99
A=\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+....+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}}\)
B=\(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}}\)
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
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
Tìm tỉ số phần trăm của A và B biết:
\(A=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+.....+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}}\) \(B=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}}\)
cho \(M=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\) và \(N=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
Tính tỉ số M với N
\(\frac{\frac{1}{99}+\frac{2}{98}+....+\frac{98}{2}+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+...+\frac{1}{500}}\)