Ta có : \(B=x^4-4x^3+9x^2-20x+22=\left(x^4-4x^3+4x^2\right)+\left(5x^2-20x+20\right)+2\)
\(=x^2\left(x^2-4x+4\right)+5\left(x^2-4x+4\right)+2=x^2\left(x-2\right)^2+5\left(x-2\right)^2+2\)
\(=\left(x-2\right)^2\left(x^2+5\right)+2\ge2\). Dấu đẳng thức xảy ra khi x = 2
Vậy Min B = 2 <=> x = 2
B=x4-4x3+9x2-20x+22
=(x-2)4+4(x-2)3+9(x-2)2+2
Ta thấy:
\(\hept{\begin{cases}\left(x-2\right)^4\\4\left(x-2\right)^3\\9\left(x-2\right)^2\end{cases}}\ge0\)
\(\Rightarrow\left(x-2\right)^4+4\left(x-2\right)^3+9\left(x-2\right)^2\ge0\)
\(\Rightarrow\left(x-2\right)^4+4\left(x-2\right)^3+9\left(x-2\right)^2+2\ge0+2=2\)
\(\Rightarrow B\ge2\)
Dấu = khi (x-2)4=4(x-2)3=9(x-2)2=0 =>x=2
Vậy Bmin=2 <=>x=2
