\(P=\left(a^2+b^2-c^2\right)\cdot x^2-4abx+a^2+b^2-c^2=0\)
Xét \(\Delta=\left(4ab\right)^2-4\left(a^2+b^2-c^2\right)\cdot\left(a^2+b^2-c^2\right)\)
\(=\left(4ab\right)^2-\left[2\left(a^2+b^2-c^2\right)\right]^2\)
\(=\left[4ab-2\left(a^2+b^2-c^2\right)\right]\left[4ab+2\left(a^2+b^2-c^2\right)\right]\)
\(=4\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(a+b+c\right)\)
Do \(a,b,c\) là độ dài 3 cạnh tam giác nên \(\left\{{}\begin{matrix}a+b-c>0\\a-b+c>0\\b+c-a>0\\a+b+c>0\end{matrix}\right.\)
Do đó \(\Delta>0\) nên pt luôn có nghiệm.
\(P=(a^2+b^2-c^2)x^2-4abx+a^2+b^2-c^2=0 \)