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\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)
\(\Leftrightarrow a^2b+a^2c+b^2a+b^2c+abc+abc+bc^2+ac^2=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\Leftrightarrow...\)
\(P=0\)
Ta có: \(a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự: \(\left\{{}\begin{matrix}b^2+1=\left(a+b\right)\left(b+c\right)\\c^2+1=\left(c+a\right)\left(b+c\right)\end{matrix}\right.\)
=> \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
Mặt khác: \(a+b+c-abc=a\left(1-bc\right)+b+c\)
\(=a\left(ab+ca\right)+b+c\) (Vì ab+bc+ca=1)
\(=\left(a^2+1\right)\left(b+c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (Vì \(a^2+1=\left(a+b\right)\left(c+a\right)\))
\(T=1\)
\(a+b+c=7\Rightarrow a+b+c-1=6\)
Ta có:\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow49=23+2\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca=13\)
Lại có \(ab+c-6=ab+c-\left(a+b+c-1\right)=ab-a-b+1=\left(a-1\right)\left(b-1\right)\)
Tương tự \(bc+a-6=\left(b-1\right)\left(c-1\right)\)
\(ca+b-6=\left(c-1\right)\left(a-1\right)\)
\(\Rightarrow A=\frac{1}{\left(a-1\right)\left(b-1\right)}+\frac{1}{\left(b-1\right)\left(c-1\right)}+\frac{1}{\left(c-1\right)\left(a-1\right)}\)
\(=\frac{c-1+a-1+b-1}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=\frac{a+b+c-3}{abc-\left(ab+ac+bc\right)+\left(a+b+c\right)-1}\)
\(=\frac{7-3}{3-13+7-1}=-1\)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow a+b=0\Leftrightarrow a=-b\)
\(\Rightarrow a^{23}+b^{23}=-b^{23}+b^{23}=0\)
Vậy \(\left(a^{23}+b^{23}\right)\left(a^{1995}+c^{1995}\right)=0\)