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\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
Bài 1b:
\(\frac31\) + \(\frac33\) + \(\frac36\) + \(\frac{3}{10}\) + ...+\(\frac{3}{x\left(x+1\right):2}\) = \(\frac{2015}{336}\)
3.(\(\frac11+\frac13+\frac16+\frac{1}{10}+\cdots+\frac{1}{x\left(x+1):2\right.})\) = \(\frac{2015}{336}\)
3.2(\(\frac12+\frac16+\frac{1}{12}+\frac{1}{20}+\cdots+\frac{1}{x\left(x+1\right)})=\) \(\frac{2015}{336}\)
6.(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\cdots+\frac{1}{x.\left(x+1\right)})\) = \(\frac{2015}{336}\)
6.(\(\frac11-\frac12\) + \(\frac12\)-\(\frac14\) +...+ \(\frac{1}{x}\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)
6.(\(\frac11\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)
1 - \(\frac{1}{x+1}\) = \(\frac{2015}{336}\) : 6
1 - \(\frac{1}{x+1}\) = \(\frac{2015}{2016}\)
\(\frac{1}{x+1}\) = 1 - \(\frac{2015}{2016}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2016}\)
\(x+1\) = 2016
\(x\) = 2016 - 1
\(x\) = 2015
Bài 2:
A = \(\frac{6n+1}{4n+3}\) (n ∈ Z\(^{-}\))
A ∈ Z khi và chỉ khi:
(6n + 1) ⋮ (4n + 3)
(12n + 2) ⋮ (4n + 3)
[3(4n + 3) - 7] ⋮ (4n + 3)
7 ⋮ (4n + 3)
(4n + 3) ∈ Ư(7) = {-7; -1; 1; 7}
n ∈ {- 5/2; -1; - 1/2; 1}
Nếu n = - 1 thì A = (-6 + 1)/(-4 + 3) = 5 (loại)
Nếu n = 1 thì: A = (6 + 1).(4+3) = 1 (loại)
Không có giá trị nào thỏa mãn đề bài hay n ∈ ∅
a, (n+1)(n+3) là SNT <=> 1 ts = 1; ts còn lại là SNT.
TH1: n+1=1 => n=0 => n+3=3 (t/m)
TH2: n+3=1 => n=-2 => n+1=-1 (không t/m)
=> n=0.
b, A không tối giản => ƯCLN(n+3;n-5) >1
=> ƯCLN(8;n-5) >1 => n-5 chẵn => n lẻ.
Giả sử \(a< b< c\)thì \(a\ge2\)\(;\)\(b\ge3\)\(;\)\(c\ge5\)
Ta có:
\(\frac{1}{\left[a,b\right]}=\frac{1}{ab}\le\frac{1}{6}\)\(;\)\(\frac{1}{\left[b,c\right]}=\frac{1}{bc}\le\frac{1}{15}\)\(;\)\(\frac{1}{\left[c,a\right]}=\frac{1}{ca}\le\frac{1}{10}\)
Do đó: \(\frac{1}{\left[a,b\right]}+\frac{1}{\left[b,c\right]}+\frac{1}{\left[c,a\right]}\le\)\(\frac{1}{6}+\frac{1}{15}+\frac{1}{10}=\frac{1}{3}\)
\(\Rightarrow\)\(\frac{1}{\left[a,b\right]}+\frac{1}{\left[b,c\right]}+\frac{1}{\left[c,a\right]}\le\)\(\frac{1}{3}\)\(\rightarrowĐPCM\)