\(Q\left(x+1\right)^2=2x^2+2\)
\(\Leftrightarrow Q\left(x^2+2x+1\right)-2x^2-2=0\)
\(\Leftrightarrow x^2\left(Q-2\right)+2Qx+\left(Q-2\right)=0\)
\(\Delta'=Q^2-\left(Q-2\right)^2\ge0\Leftrightarrow\left(Q+Q-2\right)\left(Q-Q+2\right)\ge0\Leftrightarrow2\left(2Q-2\right)\ge0\Leftrightarrow Q\ge1\)
Sau khi đoán được cực trị bằng 1 ta tách như sau:
\(\dfrac{2x^2+2}{\left(x+1\right)^2}=\dfrac{x^2+2x+1+x^2+1-2x}{\left(x+1\right)^2}=1+\dfrac{\left(x-1\right)^2}{\left(x+1\right)^2}\ge1\)
\("="\Leftrightarrow x=1\)
\(Q=\dfrac{2x^2+2}{\left(x+1\right)^2}=\dfrac{x^2-2x+1+x^2+2x+1}{\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2}{\left(x+1\right)^2}+1\ge1\)
Vậy GTNN của Q = 1 khi x = 1
