\(a,P=2x^2+10x+25=2\left(x^2+5x+\dfrac{25}{2}\right)\)
\(=2\left[x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{25}{4}+\dfrac{25}{2}\right]\)
\(=2\left[\left(x+\dfrac{5}{2}\right)^2+\dfrac{25}{4}\right]\)
\(=2\left(x+\dfrac{5}{2}\right)^2+\dfrac{25}{2}\ge\dfrac{25}{2}>0\) với mọi x (dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{5}{2}\))
Vậy \(P_{min}=\dfrac{25}{2}\) tại \(x=-\dfrac{5}{2}\)
\(b,Q=5x-x^2=-\left(x^2-5x\right)\)
\(=-\left[x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{25}{4}\right]\)
\(=-\left[\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\right]\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}< 0\) với mọi x (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\))
Vậy \(Q_{max}=\dfrac{25}{4}\) tại \(x=\dfrac{5}{2}\)
C, đề sai