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3/ Ta có \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=\left[ab\left(a+b\right)+abc\right]+\left[bc\left(b+c\right)+abc\right]+\left[ca\left(c+a\right)+ca\right]-abc\)
\(=\left(a+b+c\right)ab+\left(a+b+c\right)bc+\left(a+b+c\right)ca-abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)= -abc
Suy ra \(P=\frac{-abc}{abc}=-1\)
Vậy..
Vì \(c^2+2\left(ab-ac-bc\right)=0\) nên :
\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+\left(a-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}{b^2+\left(b-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}\)
\(=\frac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}=\frac{\left(a-c\right)^2+b\left(a-c\right)}{\left(b-c\right)^2+a\left(b-c\right)}\)
\(=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}=\frac{a-c}{b-c}\) \(\left(b\ne c,a+b\ne0\right)\)
\(P=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ca+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(=\frac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)
Do \(c^2+2\left(ab-ac-bc\right)=0\Leftrightarrow-c^2=2\left(ab-ac-bc\right)\)
Ta có; \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+c^2-c^2+\left(a-c\right)^2}{b^2+c^2-c^2+\left(b-c\right)^2}=\frac{a^2+c^2+2\left(ab-ac-bc\right)+\left(a-c\right)^2}{b^2+c^2+2\left(ab-ac-bc\right)+\left(b-c\right)^2}\)
\(=\frac{2\left(a-c\right)^2+2\left(ab-bc\right)}{2\left(b-c\right)^2+2\left(ab-ac\right)}=\frac{2\left(a-c\right)^2+2b\left(a-c\right)}{2\left(b-c\right)^2+2a\left(b-c\right)}=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}\)
\(=\frac{a-c}{b-c}\) (đpcm)
\(a+b+c=1\Rightarrow\hept{\begin{cases}ab+c=ab+c\left(a+b+c\right)\\bc+a=bc+a\left(a+b+c\right)\\ca+b=ca+b\left(a+b+c\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=ab+ca+bc+c^2\\bc+a=bc+a^2+ab+ac\\ca+b=ca+ab+b^2+bc\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=\left(b+c\right)\left(a+c\right)\\bc+a=\left(a+c\right)\left(a+b\right)\\ca+b=\left(b+c\right)\left(a+b\right)\end{cases}}\)
\(\Rightarrow P=\frac{\left(b+c\right)\left(a+c\right)}{\left(a+b\right)^2}.\frac{\left(a+c\right)\left(a+b\right)}{\left(b+c\right)^2}.\frac{\left(b+c\right)\left(a+b\right)}{\left(c+a\right)^2}=1\)
\(P=\frac{ab+c}{\left(a+b\right)^2}.\frac{bc+a}{\left(b+c\right)^2}.\frac{ca+b}{\left(c+a\right)^2}\)
\(=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ca+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(=\frac{\left(c+a\right)\left(c+b\right)}{\left(a+b\right)^2}.\frac{\left(a+b\right)\left(a+c\right)}{\left(b+c\right)^2}.\frac{\left(b+a\right)\left(b+c\right)}{\left(c+a\right)^2}=1\)