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\(\frac{9.11+19.9}{27.5-27.15}=\frac{9.\left(11+19\right)}{27.\left(5-15\right)}=\frac{9.30}{3.9.\left(-10\right)}=\frac{30}{3.\left(-10\right)}=\frac{30}{-30}=-1\)
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\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+...+\frac{99}{101.200}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=2.\frac{39}{200}=\frac{39}{100}\)
\(E=\frac{6}{5.8}+\frac{22}{8.19}+\frac{24}{19.31}+\frac{140}{31.101}+\frac{198}{101.200}\)
\(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=2.\frac{39}{200}\)
\(=\frac{39}{100}\)
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\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+3}=-\frac{3}{10}\)
\(\Leftrightarrow1\cdot10=-3\left(x+3\right)\)
\(\Leftrightarrow10=-3x-9\)
\(\Leftrightarrow10+9=-3x\)
\(\Leftrightarrow19=-3x\)
\(\Leftrightarrow x=-\frac{19}{3}\)
Đề sai à -.-
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
=> \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}:\frac{1}{3}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\cdot3=\frac{1}{2}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\)
=> \(10=-3\left(x+3\right)\)
=> 10 = -9x - 9
=> 10 + 9x + 9 = 0
=> 19 + 9x = 0
=> 9x = -19
=> x = -19/9
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a) Ý bạn là: \(S_1=\frac{3}{4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)đúng không?
\(S_1=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(S_1=1-\frac{1}{43}< 1\left(đpcm\right)\)
b) \(S_2=\frac{6}{2\cdot5}+\frac{6}{5.8}+\frac{6}{8\cdot11}+...+\frac{6}{29\cdot32}\)
=>\(\frac{S_2}{2}=\frac{3}{2\cdot5}+\frac{3}{5.8}+\frac{3}{8\cdot11}+...+\frac{3}{29\cdot32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{32}=\frac{16}{32}-\frac{1}{32}=\frac{15}{32}\)
=>\(S_2=\frac{15}{32}\cdot2=\frac{15}{16}< 1\left(đpcm\right)\)
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đề sai thì phải
\(A=\frac{10}{2\cdot12}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{2\cdot3}+\frac{5}{12\cdot17}+\frac{6}{17\cdot23}+\frac{7}{23\cdot30}\)
\(A=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{2}-\frac{1}{3}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(A=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}\)
\(A=\frac{101}{120}\)
\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+...+\frac{7}{23.30}\)
\(=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}=1-\frac{19}{120}=\frac{101}{120}\)
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các p/s lập được từ đẳng thức -2.21=7.-6là
\(\frac{-2}{7}=\frac{-6}{21};\frac{-2}{-6}=\frac{7}{21};\frac{7}{-2}=\frac{21}{-6};\frac{-6}{-2}=\frac{21}{7}\)\(\frac{21}{-7}\)