Question

tìm GTNN của bt

(x+2)(x-1)(x+5)(x+8)

Answer 1

\(A=\left(x+2\right)\left(x-1\right)\left(x+5\right)\left(x+8\right)\)

\(A=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x-1\right)\left(x+8\right)\right]\)

\(A=\left(x^2+7x+10\right)\left(x^2+7x-8\right)\)

Đặt \(t=x^2+7x+1\)ta có :

\(A=\left(t+9\right)\left(t-9\right)\)

\(A=t^2-9^2=t^2-81\ge-81\)

Dấu "=" xảy ra \(\Leftrightarrow x^2+7x+1=0\)