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a,=1/3-1/4+1/4-1/5+.............+1/20-1/21
=1/3-1/21
=2/7
b,=1/2(1/4.6+1/6.8+............+1/30.32)
=1/2(1/4-1/6+1/5+1/8+.............+1/30-1/32)
=1/2(1/4-1/32)
=1/2.7/32
=1/64
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a, 1/1.2+1/2.3+1/3.4+...+1/999.1000
= 1/1-1/2+1/2-1/3+1/3-1/4+....+1/999-1/1000
= 1/1-1/1000
= 999/1000
b, 1/2.4+1/4.6+1/6.8+1/8.10
= 1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10
= 1/2-1/10
= 4/10 =2/5
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\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(2A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(2A=\frac{1}{1}-\frac{1}{100}\)
\(2A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:2\Rightarrow A=\frac{99}{200}\)
Câu B và C làm tương tự.
bạn Nhi làm sai rồi
\(\frac{2}{2\cdot3}\) sao có thể bằng \(\frac{1}{2}-\frac{1}{3}\) được
\(\frac{1}{2\cdot3}\) mới bằng \(\frac{1}{2}-\frac{1}{3}\)
kết quả là : \(\frac{49}{100}\)
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\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)
A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)
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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2106}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)
\(A=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2014.2016}=\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1007.1008}\right)\)
=> \(B=\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{1008}\right)=\frac{1}{4}.\frac{1007}{1008}\)
=> \(B=\frac{1007}{4032}\)
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\(A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}.\frac{1009}{2020}\)
\(A=\frac{1009}{4040}\)
A=1/2.4+1/4.6+1/6.8+...+1/2018.2020
=1/2(1/2-1/4+1/4-1/6+...+1/2018-1/2020)
=1/2(1/2-1/2020)
=1/2.1009/2020
=1009/4040
a,ta thay hai so o duoi mau so la hai sop tu nhien lien tiep khoang cach chinh bang 1
ta co 1/3-1/4+1/4-1/5+.............................+1/20-1/21
ta co =1/3-1/21 vi co cac so doi to da the hien tren
=2/7
b vi khoang cach duoi mau kac tu mau la 2 con tu la 1 vay nhan 2 vao ca day so ta duoc
2/4.6+2/6.8+..............................+2/30.32
bay gio khoang cach duoi mau bang tu ta co
1/4-1/6+1/6-1/8+............................+1/30-1/32
nhu tren ta co =(1/4-1/32):2=7/64