\(\Delta=\left(b^2+c^2-a^2\right)^2-4b^2c^2\)
\(=\left(b^2+c^2-a^2\right)^2-\left(2bc\right)^2\)
\(=\left(b^2+c^2-2bc-a^2\right)\left(b^2+c^2+2bc-a^2\right)\)
\(=\left[\left(b-c\right)^2-a^2\right]\left[\left(b+c\right)^2-a^2\right]\)
\(=\left(b-c-a\right)\left(b-c+a\right)\left(b+c-a\right)\left(b+c+a\right)\)
Theo định lý tam giác, ta luôn có:
\(\left\{{}\begin{matrix}b< a+c\\a+b>c\\b+c>a\\a+b+c>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b-a-c< 0\\a+b-c>0\\b+c-a>0\\a+b+c>0\end{matrix}\right.\) \(\Rightarrow\Delta< 0\)
\(\Rightarrow\) Pt đã cho vô nghiệm
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