\(\left\{{}\begin{matrix}a^2=b^2-a\\c^2+c=a^2\end{matrix}\right.\) \(\Rightarrow c^2+c=b^2-a\Rightarrow a+c=b^2-c^2=\left(b-c\right)\left(b+c\right)\)
\(\Rightarrow b-c=\dfrac{a+c}{b+c}\)
\(\left\{{}\begin{matrix}b^2=c^2-b\\a^2+a=b^2\end{matrix}\right.\) \(\Rightarrow c^2-b=a^2+a\Rightarrow c-a=\dfrac{a+b}{a+c}\)
\(\left\{{}\begin{matrix}c^2=a^2-c\\b^2+b=c^2\end{matrix}\right.\) \(\Rightarrow b^2+b=a^2-c\Rightarrow a-b=\dfrac{b+c}{a+b}\)
Vậy \(\left(a-b\right)\left(b-c\right)\left(c-a\right)=\dfrac{\left(b+c\right)}{\left(a+b\right)}.\dfrac{\left(a+c\right)}{\left(b+c\right)}.\dfrac{\left(a+b\right)}{\left(a+c\right)}=1\) (đpcm)
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