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Giải:
Đặt \(A=\left(a+b\right)\left(b+c\right)\left(c+a\right)-2abc\)
\(=\left(a+b+c-c\right)\left(b+c\right)\left(c+a\right)-2abc\)
\(= (a+b+c)(b+c)(c+a) - c(b+c)(c+a) - 2abc \)
\(= (a+b+c)(b+c)(c+a) - c(bc + ab + c² + ca + ab) \)
\(= (a+b+c)(b+c)(c+a) - c[c(b+c+a) + 3ab] \)
\(= (a+b+c)(b+c)(c+a) - c²(a+b+c) - 3abc \)
Vì \(\left(a+b+c\right)⋮6\Rightarrow\left(a+b+c\right)⋮2\Rightarrow a+b+c\) là 1 số chẵn
\(\Rightarrow\) Trong 3 số \(a,b,c\) phải có ít nhất 1 số chẵn (vì 3 số lẻ \(\Rightarrow a+b+c\) lẻ)
\(\Rightarrow abc⋮2\Rightarrow3abc⋮6\Rightarrow A⋮6\rightarrow\) Đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
đồ vô ơn.tao đã giải cho câu a rùi mà ko tick thi thui.xéo
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a) Ta có:
\(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
\(=-5n\)
Vì \(-5n⋮5\) với n thuộc Z
\(\Rightarrow n\left(2n-3\right)-2n\left(n+1\right)⋮5\) với n thuộc Z
b) Ta có:
\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)
\(=5n^2+5n\)
\(=5\left(n^2+n\right)\)
Vì \(5\left(n^2+n\right)⋮5\)
\(\Rightarrow\left(n^2+3n-1\right)\left(n+2\right)-n^3+2⋮5\)
c) Ta có:
\(\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)
\(=\left(xy+1-2\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)
\(=\left(xy+1\right)\left(x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)
\(=\left(xy+1\right)\left(x^{2003}+y^{2003}-x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)\)
\(=2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)\)
Vì \(2\left(xy+1\right)y^{2003}⋮2\)
\(2\left(x^{2003}+y^{2003}\right)⋮2\)
\(\Rightarrow2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)⋮2\)
\(\Rightarrow\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)⋮2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)