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Bài 1 :
a) =) \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)= \(1-\frac{1}{101}=\frac{100}{101}\)
b) =) \(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
=) \(\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)( theo phần a)
Bài 2 :
-Gọi d là UCLN \(\left(2n+1;3n+2\right)\)( d \(\in N\)* )
(=) \(2n+1⋮d\left(=\right)3.\left(2n+1\right)⋮d\)
(=) \(6n+3⋮d\)
và \(3n+2⋮d\left(=\right)2.\left(3n+2\right)⋮d\)
(=) \(6n+4⋮d\)
(=) \(\left(6n+4\right)-\left(6n+3\right)⋮d\)
(=) \(6n+4-6n-3⋮d\)
(=) \(1⋮d\left(=\right)d\in UC\left(1\right)\)(=) d = { 1;-1}
Vì d là UCLN\(\left(2n+1;3n+2\right)\)(=) \(d=1\)(=) \(\frac{2n+1}{3n+2}\)là phân số tối giản ( đpcm )
Bài 3 :
-Để A \(\in Z\)(=) \(n+2⋮n-5\)
Vì \(n-5⋮n-5\)
(=) \(\left(n+2\right)-\left(n-5\right)⋮n-5\)
(=) \(n+2-n+5⋮n-5\)
(=) \(7⋮n-5\)(=) \(n-5\in UC\left(7\right)\)= { 1;-1;7;-7}
(=) n = { 6;4;12;-2}
Vậy n = {6;4;12;-2} thì A \(\in Z\)
Bài 4:
A = \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
= \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
= \(10101.\left(\frac{1}{111111}+\frac{5}{222222}\right)\)= \(10101.\left(\frac{2}{222222}+\frac{5}{222222}\right)\)
= \(10101.\frac{7}{222222}\)( không cần rút gọn \(\frac{7}{222222}\))
= \(\frac{7}{22}\)
\(2S=\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{97}-\frac{2}{99}\)
\(2S=2-\frac{2}{99}\)
\(2S=\frac{196}{99}\)
\(S=\frac{196}{99}\cdot\frac{1}{2}=\frac{98}{99}\)
Ta có: S=2/1.3+2/3.5+...+2/97.99
S= 2/2.(1-1/3+1/3-1/5+...+1/97-1/99)
S= 1-1/99=98/99
a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
TA CÓ:\(1\cdot3\cdot....\cdot99=\frac{\left(1\cdot3\cdot...\cdot99\right)\left(2\cdot4\cdot...\cdot100\right)}{2\cdot4....\cdot100}=\frac{1\cdot2\cdot3\cdot....\cdot100}{2\cdot2\cdot2\cdot...\cdot2\left(50\right)\cdot1\cdot2\cdot3\cdot..\cdot50}\)
\(=\frac{51\cdot52\cdot...\cdot100}{2\cdot2\cdot2\cdot...\cdot2}=\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot...\cdot\frac{100}{2}\)(ĐPCM)
n=\(\frac{2}{3}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\times\frac{98}{99}\)
n=\(\frac{196}{297}\)
Câu \(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{2}{99.100}\)Bạn viết \(\frac{3}{99.100}=\frac{2}{99.100}\)mik sửa lại nhé.
\(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\)
\(M=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{100-99}{99.100}\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}\)
\(N=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{99-97}{97.99}\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{3}{2}.\frac{98}{99}=\frac{49}{33}\)
Ta thấy : \(\frac{297}{200}>\frac{49}{33}\Rightarrow M>N\)
A = 2/3*5 + 2/5*7 + 2/7*9 + ... + 2/97*99
A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99
A = 1/3 - 1/99
A = 32/99
\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(A=\frac{1}{3}-\frac{1}{99}\)
\(A=\frac{32}{99}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
\(\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-x\)\(=\frac{-100}{99}\)
\(\left(1-\frac{1}{99}\right)-x=\frac{-100}{99}\)
\(\frac{98}{99}-x=\frac{-100}{99}\)
\(x=\frac{98}{99}-\left(-\frac{100}{99}\right)\)
\(x=\frac{198}{99}=2\)
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