Ta có: \(\left|x^2-4x+21\right|+\left|x^2-6x+10\right|\ge\left|x^2-4x+21+x^2-6x+10\right|\)
\(\Leftrightarrow\left|x^2-4x+21\right|+\left|x^2-6x+10\right|\ge\left|2x^2-10x+31\right|\)
\(\Leftrightarrow\left|\left(x^2-4x+21\right)\cdot\left(x^2-6x+10\right)\right|\ge\left(x^2-4x+21\right)\cdot\left(x^2-6x+10\right)\)
Dấu '=' xảy ra khi: \(\left(x^2-4x+21\right)\left(x^2-6x+10\right)\ge0\)
mà \(x^2-4x+21>0\forall x\)
và \(x^2-6x+10>0\forall x\)
nên \(\left(x^2-4x+21\right)\left(x^2-6x+10\right)>0\)
\(x^2-4x+21=\left(x-2\right)^2+17>0;\forall x\)
\(\Rightarrow\left|x^2-4x+21\right|=x^2-4x+21\)
Tương tự: \(x^2-6x+10=\left(x-3\right)^2+1>0;\forall x\)
\(\Rightarrow\left|x^2-6x+10\right|=x^2-6x+10\)
Do đó \(P=x^2-4x+21+x^2-6x+10\)
\(=2x^2-10x+31=2\left(x-\frac{5}{2}\right)^2+\frac{37}{2}\ge\frac{37}{2}\)
\(P_{min}=\frac{37}{2}\) khi \(x=\frac{5}{2}\)
