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\(=\frac{3}{3.13}+\frac{3}{13.23}+...+\frac{3}{1993.2003}\)
\(=\frac{1}{10}.\left(1-\frac{3}{13}+\frac{3}{13}-\frac{3}{23}+...+\frac{3}{1993}-\frac{3}{2003}\right)\)
\(=\frac{1}{10}.\left(1-\frac{3}{2003}\right)\)
\(=\frac{1}{10}.\frac{2000}{2003}\)
\(=\frac{200}{2003}\)
Đặt \(A=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(\Rightarrow A=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(\Rightarrow A=3\left(\frac{1}{3.13}+\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+...+\frac{10}{1993.2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)\)
\(\Rightarrow A=\frac{3}{10}.\left(\frac{2003}{6009}-\frac{3}{6009}\right)\)
\(\Rightarrow A=\frac{3}{10}.\frac{2000}{6009}\)
\(\Rightarrow A=\frac{200}{2003}\)
\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{2.1}\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{1.2}+\frac{1}{3.2}+...+\frac{1}{2001.2002}\right)\)
\(=\frac{1}{2003.2002}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\right)\)
\(=\frac{1}{2003.2002}-\left(1-\frac{1}{2002}\right)\)
\(=\frac{1}{2003.2002}-\frac{2001}{2002}\)
Dễ mà bạn:
a) Cho A=1/100-1/100.99-1/99.98-...-1/3.2-1/2.1
\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=-\frac{49}{50}\)
b) \(-15,5.20,8+3,5.9,2-15,5.9,2+3,5.20,8=-15,5.\left(20,8+9,2\right)+3,5\left(9,2+20,8\right)\)
\(=-15,5.30+3,5.30=30\left(3,5-15,5\right)=30.\left(-12\right)=-360\)
A = 1/99 - 1/99.98 - 1/98.97 - ............... - 1/3.2 - 1/2.1
\(A=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
đặt \(B=\frac{1}{99.98}+\frac{1}{97.87}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(B=1-\frac{1}{99}\)
\(B=\frac{98}{99}\)
\(\Rightarrow A=\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)
e) \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\frac{99}{100}\)
\(=\frac{-98}{100}\)\
\(=\frac{-49}{50}\)
g)-15,5 . 20,8 + 3,5.9,2 - 15,5.9,2 + 3,5.20,8
=20,8.(-15.5+3,5)+9,2(-15.5+3.5)
=(-15.5+3.5)(20.8+9.2)
=(-12).30
=-360
a) \(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
Đặt A = \(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
A = \(1-\frac{1}{99}\)
A = \(\frac{98}{99}\)
Thay A vào ta được :
\(\frac{1}{100.99}-\frac{98}{99}=\frac{1}{9900}-\frac{98}{99}=\frac{-9799}{9900}\)
b) \(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-3,6.21\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
Ta thấy biểu thức trong ngoặc thứ ba của tử số có kết quả bằng 0
\(\Rightarrow\)Phân số ấy có kết quả bằng 0
\(E=\frac{2}{3.5}+\frac{7}{5.12}+\frac{9}{4.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{12}+\frac{27}{12.39}=\frac{1}{3}-\frac{1}{12}+\frac{1}{12}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)