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Bài 1:
a) \(\left(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}\right)-\left(-\frac{1}{511}-\frac{1}{2}+\frac{5}{19}\right)\)
= \(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}+\frac{1}{511}+\frac{1}{2}-\frac{5}{19}\)
= \(\left(\frac{5}{19}-\frac{5}{19}\right)+\left(\frac{1}{511}-\frac{1}{511}\right)+\left(\frac{7}{12}+\frac{1}{2}\right)\)
= 0 + 0 + \(\frac{13}{12}\)
= \(\frac{13}{12}\).
b) \(-\left(\frac{13}{25}-\frac{4}{191}+\frac{2}{51}\right)+\left(\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\right)\)
= \(-\frac{13}{25}+\frac{4}{191}-\frac{2}{51}+\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\)
= \(\left(-\frac{13}{25}+\frac{3}{5}\right)+\left(\frac{4}{191}+\frac{4}{191}\right)+\left(\frac{2}{51}-\frac{2}{51}\right)\)
= \(\frac{2}{25}+\frac{8}{191}+0\)
= \(\frac{582}{4775}\).
Mình chỉ làm câu a) và câu b) thôi nhé.
Chúc bạn học tốt!
a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
\(A=1-\frac{1}{99}\)
\(A=\frac{98}{99}\)
thay A vào, ta được :
\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)
b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)
\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)
đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(A=2.\left(1-\frac{1}{99}\right)\)
\(A=2.\frac{98}{99}\)
\(A=\frac{196}{99}\)
Thay A vào, ta được :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{35.37}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{35}-\frac{1}{37}\)
\(=\frac{1}{3}-\frac{1}{37}=\frac{34}{111}\)
c) \(\frac{7}{7.9}+\frac{7}{9.11}+\frac{7}{11.13}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{101}\right)=\frac{7}{2}\cdot\frac{94}{707}=\frac{47}{101}\)
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{999}{2000}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2000}\)
\(\Leftrightarrow x+1=2000\)\(\Leftrightarrow x=1999\)
Vậy \(x=1999\)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{15.2}{93}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow2x=90\)\(\Leftrightarrow x=45\)
Vậy \(x=45\)
\(A=1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-...-\frac{2}{63.65}\)
\(A=1-\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{63-65}\right)\)
\(A=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{63}-\frac{1}{65}\right)\)
\(A=1-\left(\frac{1}{3}-\frac{1}{65}\right)\)
\(A=1-\frac{62}{195}\)
\(A=\frac{133}{195}\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)
\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)
\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)
\(a,\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x^2=8^2\)
\(\Leftrightarrow x=\pm8\)
\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Leftrightarrow x=15\)
Vậy x = 15
Bài cuối tương tự
\(a)\) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{197.198}+\frac{1}{198.199}\right)\)
\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-\left(1-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-1+\frac{1}{199}\)
\(A=\frac{-197}{199}\)
Chúc bạn học tốt ~
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