\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)
A=\(\frac{3}{1\cdot4}\)+\(\frac{3}{4\cdot7}\)+\(\frac{3}{7\cdot10}\)+\(\frac{3}{10\cdot13}\)+\(\frac{3}{13\cdot16}\)
\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(A=1-\frac{1}{16}=\frac{15}{16}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}=1-\frac{1}{16}=\frac{15}{16}\)
A=\(\frac{^{3^2}}{1\cdot4}\)+ \(\frac{3^2}{4\cdot7}\)+ \(\frac{3^2}{7\cdot10}\)+ \(\frac{3^2}{10\cdot13}\)+\(\frac{3^2}{13\cdot16}\)+......+ \(\frac{3^2}{97\cdot100}\)
A = \(\frac{3^2}{1\cdot4}+\frac{3^2}{4\cdot7}+\frac{3^2}{7\cdot10}+\frac{3^2}{10\cdot13}+\frac{3^2}{13\cdot16}+...+\frac{3^2}{97\cdot100}\)
A : 3 = \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+...+\frac{3}{97\cdot100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{97}-\frac{1}{100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{100}\)
A : 3 = \(\frac{99}{100}\)
A = \(\frac{297}{100}\)
B=\(\frac{5}{1\cdot4}\)+\(\frac{5}{4\cdot7}\)+\(\frac{5}{7\cdot10}\)+\(\frac{5}{10\cdot13}\)+\(\frac{5}{13\cdot16}\)
Ta có :
\(B=\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+\frac{5}{13.16}\)
\(\frac{3}{5}B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(\frac{3}{5}B=1-\frac{1}{16}\)
\(B=\frac{15}{16}:\frac{3}{5}\)
\(B=\frac{25}{16}\)
Ủng hộ mk nha !!! ^_^
\(B=\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+\frac{5}{13.16}\)
\(\frac{3}{5}B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(\frac{3}{5}B=1-\frac{1}{16}\)
\(B=\frac{15}{16}:\frac{3}{5}\)
\(B=\frac{25}{16}\)
Chứng tỏ rằng :
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...\frac{1}{2010\cdot2013}<\frac{1}{3}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{2010.2013}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2010}-\frac{1}{2013}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{2013}\right)=\frac{1}{3}.\frac{2012}{2013}
Tìm a biết :
A=\(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+..........+\frac{3}{94\cdot97}+\frac{3}{97\cdot100}\)=??????????????????
A = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{94.97}+\frac{3}{97.100}\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{100}\)
\(\Rightarrow A=\frac{24}{100}=\frac{6}{25}\)
A=3/4-3/7+3/7-3/10+3/10-3/13+.........+3/94-3/97+3/97-3/100
Bây giờ loại các phân số giống nhau
A=3/4-3/100
A=18/25
\(\frac{5}{1\cdot4\cdot7}+\frac{5}{4\cdot7\cdot10}+\frac{5}{7\cdot10\cdot13}+.....+\frac{5}{31\cdot34\cdot37}\)Tính nhanh ( giải kiểu lớp 5 và dấu . là nhân)
THANKS nhiều
thực hiện phếp tính
\(\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot3+9\cdot12+12\cdot16+15\cdot20}\)
cảm ơn kết quả thì mik b òi nhưng mik cần cách làm
\(\frac{7}{3\cdot4}-\frac{9}{4\cdot5}+\frac{11}{5\cdot6}-\frac{13}{6\cdot7}+\frac{15}{7\cdot8}-\frac{17}{8\cdot9}+\frac{19}{9\cdot10}\)
ta có:(3+4)/4=3/(3*4)+4/(3*4) =1/4+1/3 chứng minh tương tự,cộng vế với vế, ta được kết quả là 13/30
\(\frac{2}{4\cdot7}-\frac{2}{5\cdot9}+\frac{2}{7\cdot10}-\frac{2}{9\cdot13}+\frac{2}{10\cdot13}-\frac{2}{13\cdot17}+...+\frac{2}{301\cdot304}-\frac{2}{401\cdot405}CM:tich,tren,< \frac{1}{15}\)