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a) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\frac{102}{103}\)
\(=\frac{34}{103}\)
b) \(\frac{1}{2000.1999}-\frac{1}{1999.1998}-\frac{1}{1998.1997}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{2000.1999}-\left(\frac{1}{1999.1998}+\frac{1}{1998.1997}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)(*)
Đặt biểu thức trong ngoặc là A ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1997.1998}+\frac{1}{1998.1999}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1997}-\frac{1}{1998}+\frac{1}{1998}-\frac{1}{1999}\)
\(A=1-\frac{1}{1999}\)
\(A=\frac{1998}{1999}\)
Thay vào biểu thức (*) ta có :
\(\frac{1}{2000.1999}-\frac{1998}{1999}\)
\(=\frac{1}{3998000}-\frac{1998}{1999}\)
\(=\frac{-3995999}{3998000}\)
c) \(\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(=\frac{-1}{1.3}+\frac{-1}{3.5}+\frac{-1}{5.7}+\frac{-1}{7.9}+...+\frac{-1}{99.101}\)
\(=\frac{-1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(=\frac{-1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{-1}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{-1}{2}.\frac{100}{101}\)
\(=\frac{-50}{101}\)
_Chúc bạn học tốt_
\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99\cdot101}\right)\)
\(2A=-\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{99}-\frac{2}{101}\right)\)
\(2A=-\left(2-\frac{2}{101}\right)\)
\(2A=-\frac{200}{101}\)
\(\Rightarrow A=-\frac{100}{101}\)
Đặt biểu thức trên là A, ta có:
\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(\Rightarrow A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(\Rightarrow A=-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(\Rightarrow2A=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow2A=-\left(1-\frac{1}{101}\right)\)
\(\Rightarrow2A=-\frac{100}{101}\)
\(\Rightarrow A=-\frac{100}{101}\div2=-\frac{50}{101}\)
B. 1/3 - 1/3 - 3/5 +3/5 + 5/7 - 5/7 + 9/11 - 9/11 -11/13 + 11/ 13 + 7/9 + 13/15
= 0 -0-0-0-0+7/9 +13/15
= 74/45
Nhiều thế :( Làm 1,2 câu thôi nhé
a) \(\frac{1}{3}+\frac{1}{4}=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\) (bị mất nét nhưng vẫn nhìn ra là số 12 nhỉ?)
b) \(\frac{-2}{5}+\frac{7}{21}=\frac{-42}{105}+\frac{35}{105}=\frac{-7}{105}=\frac{-1}{15}\)
a) \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)
\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{19}-\frac{1}{19}+\frac{1}{23}-\frac{1}{23}+\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)
\(=\frac{1}{2}-\frac{2}{27}\)
\(=\frac{23}{54}\)
b) \(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+\frac{1}{20}-...-\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\frac{19}{100}\)
\(=1-\frac{19}{500}\)
\(=\frac{481}{500}\)
Đặt A =\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)\)
Đặt B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
>\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{2}-\frac{1}{101}=\frac{99}{202}\)
Khi đó A = \(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)< 99-\frac{99}{202}\approx98,5\)
=> A < 98,5 (1)
Lại có B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
Khi đó A =\(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)>99-\frac{99}{100}=98,01\)
=> A > 98,01 (2)
Từ (1)(2) => 98,01 < A < 98,5
=> A không là số nguyên
a) Ta có: \(\frac{3}{8}-\frac{1}{5}+\frac{3}{40}\)
\(=\frac{15}{40}-\frac{8}{40}+\frac{3}{40}\)
\(=\frac{10}{40}=\frac{1}{4}\)
b) Ta có: \(\frac{21}{4}\cdot\frac{3}{8}+\frac{43}{4}\cdot\frac{3}{8}-4\cdot\frac{1}{2}\)
\(=\frac{3}{8}\left(\frac{21}{4}+\frac{43}{4}\right)-2\)
\(=\frac{3}{8}\cdot16-2\)
\(=6-2=4\)
c) Ta có: \(\frac{-5}{9}+\frac{7}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{8}{15}\)
\(=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{-2}{11}\)
\(=-1+1+\frac{-2}{11}\)
\(=\frac{-2}{11}\)
d) Ta có: \(125\%\cdot\left(\frac{-1}{2}\right)^2:\left(1\frac{5}{6}-1.5\right)+2016^0\)
\(=\frac{5}{4}\cdot\frac{1}{4}:\left(\frac{11}{6}-\frac{3}{2}\right)+1\)
\(=\frac{5}{16}\cdot3+1\)
\(=\frac{15}{16}+\frac{16}{16}=\frac{31}{16}\)
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!
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 \(x=45\)
Lời giải:
Ta có: \(G=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{9999}\)
\(\Rightarrow2.G=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+.....+\frac{2}{9999}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(\Rightarrow G=\frac{50}{101}\) . Vậy: \(\\G=\frac{50}{101}\)
\(G=\frac{1}{3}+\frac{1}{15}+...+\frac{1}{9999}\)
\(\Leftrightarrow G=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{1}{99.101}\right)\)
\(\Leftrightarrow G=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Leftrightarrow G=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(\Leftrightarrow G=\frac{1}{2}.\frac{100}{101}\)
\(\Leftrightarrow G=\frac{50}{101}\)
Vậy : \(G=\frac{50}{101}\)