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Đặt \(A=\frac{3\cdot7\cdot13\cdot37\cdot39-10101}{505050-70707}\)
\(A=\frac{\left(3\cdot7\cdot13\cdot37\right)\cdot39-10101\cdot1}{50\cdot10101+7\cdot10101}\)
\(A=\frac{10101\cdot39-10101\cdot1}{10101\cdot\left(50+7\right)}\)
\(A=\frac{10101\cdot\left(39-1\right)}{10101\cdot57}\)
\(A=\frac{10101\cdot38}{10101\cdot57}\)
\(A=\frac{38}{57}=\frac{38:19}{57:19}=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3.7.13.37.19-10101}{505050+70707}\)=\(\frac{10101.39-10101}{10101.50+10101.7}\)=\(\frac{10101.\left(39-1\right)}{10101.\left(50+7\right)}\)=\(\frac{38}{57}\)=\(\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
3.7.13.37.39-10101/505050+70707
=393939-10101/505050+70707
=10101.39-10101/505050+70707
=10101(39-1)/10101.50 +70707
=10101.38/10101+70707
=38+70707=70745
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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 \(\frac{7}{222222}\))
= \(\frac{7}{22}\)
Đề : \(\frac{3\cdot7\cdot13\cdot37\cdot39-10101}{505050-70707}\)
\(\Rightarrow\frac{\left(3\cdot7\cdot13\cdot37\right)\cdot39-10101}{505050-70707}\)
\(\Rightarrow\frac{10101\cdot39-10101}{505050-70707}\)
\(\Rightarrow\frac{10101\cdot\left(39-1\right)}{505050-70707}\)
\(\Rightarrow\frac{10101\cdot38}{50\cdot10101-7\cdot10101}\)
\(\Rightarrow\frac{10101\cdot38}{10101\cdot\left(50-7\right)}\)
\(\Rightarrow\frac{10101\cdot38}{10101\cdot43}\)
\(\Rightarrow\frac{38}{43}\)