a: Để \(n^2+2n+12\) là số chính phương thì \(n^2+2n+12=k^2\left(k\in Z\right)\)

=>\(n^2+2n+1-k^2=-11\)

=>\(\left(n+1\right)^2-k^2=-11\)
=>(n+1-k)(n+1+k)=-11

=>(n+1-k;n+1+k)∈{(1;-11);(-11;1);(-1;11);(11;-1)}

TH1: n+1-k=1 và n+1+k=-11

=>n+1-k+n+1+k=1-11

=>2n+2=-10

=>2n=-12

=>n=-6(loại)

TH2: n+1-k=-11 và n+1+k=1

=>n+1-k+n+1+k=1-11

=>2n+2=-10

=>2n=-12

=>n=-6(loại)

TH3: n+1-k=-1 và n+1+k=11

=>n+1-k+n+1+k=-1+11

=>2n+2=10

=>2n=8

=>n=4(nhận)

TH4: n+1-k=11 và n+1+k=-1

=>n+1-k+n+1+k=-1+11

=>2n+2=10

=>2n=8

=>n=4(nhận)

vậy: n=4

b: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

=>\(\frac{bc+ac+ab}{abc}=0\)

=>ab+ac+bc=0

=>ab=-ac-bc; ac=-ba-bc; bc=-ab-ac

\(a^2+2bc=a^2+bc+bc\)

\(=a^2-ab-ac+bc\)

=a(a-b)-c(a-b)

=(a-b)(a-c)

\(b^2+2ac=b^2+ac+ac\)

\(=b^2-ba-bc+ac\)

=b(b-a)-c(b-a)

=(b-a)(b-c)

\(c^2+2ab\)

\(=c^2+ab+ab\)

\(=c^2-ac-bc+ab\)

=c(c-a)-b(c-a)

=(c-a)(c-b)

=(a-c)(b-c)

\(N=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)

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

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

=0