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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)
\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)
=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)
2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{a^3.b^3+b^3.c^3+c^3.a^3}{3\left(abc\right)^2}\)
\(P=\frac{\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3}{3\left(abc\right)^2}\)
\(P=\frac{0}{3\left(abc\right)^2}\)
\(P=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(a-1\right)\left(bc-b-c+1\right)\)
\(=abc-\left(ab+bc+ca\right)+a+b+c-1\)
\(=abc-abc+1-1=0\) (đpcm)