\(B=\left(x^2-x\right)\left(x^2-x-6\right)+9\)
\(=\left(x^2-x\right)^2-6\left(x^2-x\right)+9=\left(x^2-x-3\right)^2\ge0\)
Vậy GTLN của B là 0
\(B=x\left(x-1\right)\left(x+2\right)\left(x-3\right)-9\)
\(B=\left[x\left(x-1\right)\right]\left[\left(x+2\right)\left(x-3\right)\right]-9\)
\(B=\left(x^2-x\right)\left(x^2-3x+2x-6\right)-9\)
\(B=\left(x^2-x\right)\left(x^2-x-6\right)-9\)
Đặt \(a=x^2-x-3\)
\(\Leftrightarrow B=\left(a+3\right)\left(a-3\right)-9\)
\(\Leftrightarrow B=a^2-3^2-9\)
\(\Leftrightarrow B=a^2-18\)
Thay \(a=x^2-x-3\)vào B ta có :
\(B=\left(x^2-x-3\right)^2-18\le-18\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x^2-x-3=0\Leftrightarrow x^2-x=3\)
Vậy Bmin = -18 <=> x2 - x = 3
@Pham Van Hung đề là -9 mà sao ghi +9 ?
