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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 = \(\frac{297}{100}\)
\(A=\frac{3}{2}\times\left(\frac{1}{13\times11}+\frac{1}{13\times15}+\frac{1}{15\times17}+.....+\frac{1}{97\times99}\right)\)
\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+......+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{99}\right)\)
\(A=\frac{3}{2}\times\frac{8}{99}\)
\(A=\frac{4}{33}\)
b] \(\frac{A}{5}=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)
\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)
\(\frac{A}{5}=\frac{1}{31}-\frac{1}{57}\)
\(\Rightarrow A=5\left(\frac{1}{31}-\frac{1}{57}\right)=\frac{130}{1767}\)
c] Ta đặt \(\left(8n+5,6n+4\right)=d\)
\(\Rightarrow\frac{8n+5\div d}{6n+4\div d}\Rightarrow4\times\left(6n+4\right)-3\times\left(8n+5\right)=\left(24n+16\right)-\left(24n+15\right):d\)\(\Rightarrow d=1\)
Vậy \(\frac{8n+5}{6n+4}\)là phân số tối giản
\(a,5\frac{4}{7}:x=13\Leftrightarrow x=\frac{39}{7}:13\Leftrightarrow x=\frac{39}{7}.\frac{1}{13}=\frac{3}{7}\)
\(b,\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\Leftrightarrow2,8x-32=-90.\frac{2}{3}=-60\)
\(\Leftrightarrow2,8x=-60+32=-28\)
\(\Leftrightarrow x=\frac{-28}{2,8}=-10\)
d, \(7x=3,2+3x\Leftrightarrow7x-3x=3,2\Leftrightarrow4x=3,2\Leftrightarrow x=3,2:4=3,2.\frac{1}{4}=\frac{4}{5}\)
Câu c bị sai đề :\(\frac{19}{10}-1-\frac{2}{5}=\frac{1}{2}\ne1\)bạn nha.
mình lộn \(\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=\frac{13}{10}\ne1\)ms đúng nha
\(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}=\frac{1}{7}-\frac{1}{100}=\frac{93}{700}\)
\(\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{97\cdot100}\)
\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}=\frac{1}{7}-\frac{1}{100}=\frac{93}{700}\)