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TN
2
15 tháng 6 2023
=1/50-(1-1/2+1/2-1/3+...+1/49-1/50)
=1/50-1+1/50
=1/25-1=-24/25
15 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{1}{50}-\dfrac{1}{50\cdot49}-\dfrac{1}{49\cdot48}-...-\dfrac{1}{2\cdot1}\)
`=`\(\dfrac{1}{50}-\left(\dfrac{1}{50\cdot49}+\dfrac{1}{49\cdot48}+...+\dfrac{1}{2\cdot1}\right)\)
`=`\(\dfrac{1}{50}-\left(\dfrac{1}{50}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{48}+...+\dfrac{1}{2}-1\right)\)
`=`\(\dfrac{1}{50}-\left(\dfrac{1}{50}-1\right)\)
`=`\(\dfrac{1}{50}-\left(-\dfrac{49}{50}\right)\)
`= 1`
10 tháng 10 2018
A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)
12 tháng 10 2022
\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)
\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)
=1-82/84
=2/84=1/42
\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)
\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
L
30 tháng 8 2019
Ta có :
\(A=\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(A=-\left(\frac{1}{2003\cdot2002}+\frac{1}{2002\cdot2001}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(A=-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\)
\(A=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(A=-\left(1-\frac{1}{2003}\right)\)
\(A=-\frac{2002}{2003}\)
30 tháng 8 2019
\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2001.2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}.+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}\left(1+\frac{1}{2003}\right)\)
\(=-1+\frac{1}{2002}.\frac{2004}{2003}\)
\(=-1+\frac{2}{2003}\)
\(=\frac{-2003+2}{2003}\)
\(=\frac{-2001}{2003}\)
11 tháng 9 2016
link nè bạn
/hoi-dap/question/88514.html
hoặc bạn sang trang 3 của hỏi đáp toán hoc24 sẽ thấy nhé
18 tháng 11 2021
\(a,A=\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2009}+\dfrac{1}{2008}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ A=1+\dfrac{1}{2010}=\dfrac{2011}{2010}\)
\(b,B=\left(-124\right)\left(63-37\right)+\dfrac{17}{66}\left(-66\right)=-124\cdot26+17=-3224+17=-3207\)
12 tháng 7 2016
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{-49}{50}\)
15 tháng 4 2019
\(\frac{1}{\left(a+1\right)a}=\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng đẳng thức trên ta tính ĐƯỢC:
A= 1/100-(1/99-1/100+1/98-1/99+...+1/2-1/3+1/1-1/2)
=1/100-(-1/100+1)
=1/50+1=51/50
15 tháng 4 2019
\(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\frac{99}{100}\)
\(A=\frac{-98}{100}=-\frac{49}{50}\)
Ta đặt
\(A=\dfrac{1}{50}-\dfrac{1}{50\times49}-....-\dfrac{1}{2\times1}\)
\(A=\dfrac{1}{50}-\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{49\times50}\right)\)
\(A=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(A=\dfrac{1}{50}-\left(1-\dfrac{1}{50}\right)\)
\(A=\dfrac{1}{50}-\dfrac{49}{50}\)
\(A=\dfrac{-48}{50}=\dfrac{-24}{25}\)
\(=\dfrac{1}{50}-\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\right)=\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{50}\right)=\dfrac{2}{50}-1=\dfrac{1}{25}-1=-\dfrac{24}{25}\)