\(=a^5+a^3b^2+b^3a^2+b^5-\left(a+b\right)\)

\(=a^5+b^5+\left(a^3b^2+b^3a^2\right)-\left(a+b\right)\)

\(=a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)

\(=a^5+b^5+\left[\left(ab\right)^2-1\right]\left(a+b\right)\)

Mà \(ab=1\Rightarrow\left(ab\right)^2-1=1^2-1=0\)

\(\Rightarrow\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)=a^5+b^5+0=a^5+b^5\)

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Xét \(\Delta IHB\)có IA vừa là đường cao vừa là trung tuyến nên cân tại I, nên IA đồng thời là được phân giác

\(\Rightarrow\widehat{AIB}=\widehat{AIH}\)

Mà \(\widehat{AIH}=\widehat{DIC}\)( Đối đỉnh )

\(\Rightarrow\widehat{AIB}=\widehat{DIC}\)

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\(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2+c^2-\left(a+b\right)c\right]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left[a^2+b^2+2ab+c^2-ac-bc-3ab\right]=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0.2\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

TH1 : \(a+b+c=0\)

\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

\(=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\)

\(=\frac{\left(-c\right)}{b}.\frac{\left(-a\right)}{c}.\frac{\left(-b\right)}{a}=-1\)

TH2 : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)

\(\Rightarrow a-b=b-c=c-a=0\)

\(\Rightarrow a=b=c\)

\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

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\(A=\left(n-1\right)n\left(n+1\right)\left(n+2\right)-3\)

\(=\left[n\left(n+1\right)\right]\left[\left(n-1\right)\left(n+2\right)\right]-3\)

\(=\left(n^2+n\right)\left(n^2+n-2\right)-3\)

\(=\left[\left(n^2+n-1\right)+1\right]\left[\left(n^2+n-1\right)-1\right]-3\)

\(=\left(n^2+n-1\right)^2-1^2-3\)

\(=\left[\left(n^2+2.\frac{1}{2}.n+\frac{1}{4}\right)-1,25\right]^2-4\)

\(=\left[\left(n+\frac{1}{2}\right)^2-1,25\right]^2-4\ge\left(-1,25\right)^2-4=-\frac{39}{16}\)

\(\Rightarrow MinA=-\frac{39}{16}\Leftrightarrow n=-\frac{1}{2}\)

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