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\(A=1.2.3...2018\left[\left(1+\dfrac{1}{2018}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2017}\right)+...+\left(\dfrac{1}{1009}+\dfrac{1}{1010}\right)\right]\)
\(A=1.2.3...2018.2019\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\)
\(\dfrac{A}{2019}=1.2.3...2018\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\).
Rõ ràng tích 1 . 2 ... 2018 chia hết cho các tích 1 . 2018; 2 . 2017; ...; 1009 . 1010; do đó \(\dfrac{A}{2019}\) là số tự nhiên.
Vậy A chia hết cho 2019.
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)
\(\Leftrightarrow a\left(ab+ac+bc\right)+\left(b+c\right)\left(ab+ac+bc\right)-abc=0\)
\(\Leftrightarrow a\left(ab+ac+bc-bc\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow a^2\left(b+c\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow\left(a^2+ab+ac+bc\right)\left(b+c\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)
- Nếu \(a=-c\Rightarrow a^{2006}=c^{2006}\Rightarrow c^{2006}-a^{2006}=0\Rightarrow P=0\)
- Nếu \(a=-b\Rightarrow a^{2004}=b^{2004}\Rightarrow a^{2004}-b^{2004}=0\Rightarrow P=0\)
- Nếu \(b=-c\Rightarrow b^{2005}=-c^{2005}\Rightarrow b^{2005}+c^{2005}=0\Rightarrow P=0\)
Vậy \(P=0\)
Chính bài của em:
Cho \(a,b,c\ge1\). CMR: \(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)+2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}... - Hoc24
Ta có: \(16a^4+4=16a^4+2.4a^2.2+4-16a^2\)
\(=\left(4a+2\right)^2-16a^2\)
\(=\left(4a+2\right)^2-16a^2\)
\(=\left(4a^2-4a+2\right).\left(4a^2+4a+2\right)\)
\(=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( a \(\in\) N* )
Do đó: \(16a^4+4=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( * )
Thay a lần lượt bằng 1, 2, 3, ..., 2014, ta có:
\(16.1^4+4=\left[\left(2.1-1\right)^2+1\right].\left[\left(2.1+1\right)^2+1\right]=\left(1^2+1\right).\left(3^2+1\right)\)
\(16.2^4+4=\left[\left(2.2-1\right)^2+1\right].\left[\left(2.2+1\right)^2+1\right]=\left(3^2+1\right).\left(5^2+1\right)\)
\(16.3^4+4=\left[\left(2.3-1\right)^2+1\right].\left[\left(2.3+1\right)^2+1\right]=\left(5^2+1\right).\left(7^2+1\right)\)
\(16.4^4+4=\left[\left(2.4-1\right)^2+1\right].\left[\left(2.4+1\right)^2+1\right]=\left(7^2+1\right).\left(9^2+1\right)\)
\(......\)
\(16.2005^4+4=\left[\left(2.2005-1\right)^2+1\right].\left[\left(2.2005+1\right)^2+1\right]=\left(4009^2+1\right).\left(4011^2+1\right)\)
\(16.2006^4+4=\left[\left(2.2006-1\right)^2+1\right].\left[\left(2.2006+1\right)^2+1\right]=\left(4011^2+1\right).\left(4013^2+1\right)\)
Đặt \(T=\dfrac{\left(1^4+\dfrac{1}{4}\right).\left(3^4+\dfrac{1}{4}\right)...\left(2005^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right).\left(4^4+\dfrac{1}{4}\right)...\left(2006^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow T=\dfrac{16.\left(1^4+\dfrac{1}{4}\right).16\left(3^4+\dfrac{1}{4}\right)...16\left(2005^4+\dfrac{1}{4}\right)}{16.\left(2^4+\dfrac{1}{4}\right).16\left(4^4+\dfrac{1}{4}\right)...16\left(2006^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow T=\dfrac{\left(16.1^4+4\right).\left(16.3^4+4\right)...\left(16.2005^4+4\right)}{\left(16.2^4+4\right).\left(16.4^4+4\right)...\left(16.2006^4+4\right)}\)
\(\Leftrightarrow T=\dfrac{\left(1^2+1\right).\left(3^2+1\right).\left(5^2+1\right)...\left(4009^2+1\right).\left(4011^2+1\right)}{\left(3^2+1\right).\left(5^2+1\right).\left(7^2+1\right)...\left(4011^2+1\right).\left(4013^2+1\right)}\)
\(\Leftrightarrow T=\dfrac{1^2+1}{4013^2+1}\)
\(\Leftrightarrow T=\dfrac{2}{4013^2+1}\)
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{637}{1275}\)
\(\Leftrightarrow\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{1}{2}-\dfrac{637}{1275}=\dfrac{1}{2550}\)
\(\Leftrightarrow\left(n+1\right)\left(n+2\right)=2550\)
\(\Leftrightarrow n^2+3n-2548=0\)
\(\Rightarrow n=49\)
