\(f\left(x\right)=\left(25-x^2\right)\left(x^2-4x+3\right).\)
Ta có: \(25-x^2=0.\Leftrightarrow x^2=25.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\end{matrix}\right.\)
\(x^2-4x+3=0.\Leftrightarrow\left(x-3\right)\left(x-1\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=1.\end{matrix}\right.\)
Bảng xét dấu:
\(x\) \(-\infty\) -5 1 3 5 \(+\infty\)
\(25-x^2\) - 0 + | + | + 0 -
\(x^2-4x+3\) + | + 0 - 0 + | +
\(f\left(x\right)\) - 0 + 0 - 0 + 0 -
Vậy \(f\left(x\right)>0.\Leftrightarrow x\in\left(-5;1\right)\cup\left(3;5\right).\\ f\left(x\right)< 0.\Leftrightarrow x\in\left(-\infty;-5\right)\cup\left(1;3\right)\cup\left(5;+\infty\right).\\ f\left(x\right)=0.\Leftrightarrow x\in\left\{-5;1;3;5\right\}.\)
