Question

Tìm Min

D(x)=\(\frac{\left(x^2+2x+3\right)\left(x^2+2x+9\right)}{x^2+2x+1}\)

Answer 1

\(D\left(x\right)=\frac{\left(x^2+2x+3\right)\left(x^2+2x+9\right)}{x^2+2x+1}\)

Đặt \(a=x^2+2x+1=\left(x+1\right)^2\ge0\)

\(\Rightarrow\)\(D\left(x\right)=\frac{\left(a+2\right)\left(a+8\right)}{a}=\frac{a^2+10a+16}{a}\)

Áp dụng BĐT AM-GM ta có:\(a^2+16\ge2\sqrt{a^2.16}=2.4a=8a\)

\(\Rightarrow D\left(x\right)\ge\frac{8a+10a}{a}=\frac{18a}{a}=18\)

Nên minD(x)=18 đạt được khi \(a=4\Leftrightarrow\left(x+1\right)^2=4\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)