Question

Tìm max : a) \(P=\dfrac{2}{x^2+6x+12}\)

b) \(Q=3+\dfrac{4}{x^2+2x+2}\)

Answer 1

a) \(P_{max}=\left(x^2+6x+12\right)_{min}=\left[\left(x^2+2.3x+9\right)+3\right]_{min}=\left[\left(x+3\right)^2+3\right]_{min}=3\)\(P_{max}=P\left(-3\right)=\dfrac{2}{3}\)

Answer 2

\(Q_{max}=3+\left[\dfrac{4}{x^2+2x+1+1}\right]_{max}=3+\dfrac{4}{\left[\left(x+1\right)^2+1\right]_{min}=3+\dfrac{4}{\left[\left(0\right)+1\right]}=7}\)Max Q=7 khi x=-1