Bài 7 làm mẫu nha các bài còn lại tương tự :
a, \(A=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3\)
Thấy : \(\left(x-2\right)^2\ge0\forall x\in R\)
\(\Rightarrow A=\left(x-2\right)^2-3\ge-3\)
Vậy \(Min_A=-3\Leftrightarrow x=2\)
b, Ta có : \(B=4x^2+4x+11=4x^2+4x+1+10=\left(2x+1\right)^2+10\)
Thấy : \(\left(2x+1\right)^2\ge0\forall x\in R\)
\(\Rightarrow B=\left(2x+1\right)^2+10\ge10\)
Vậy \(Min_B=10\Leftrightarrow x=-\dfrac{1}{2}\)
c, Ta có : \(C=3x^2-6x-1=3x^2-6x+3-4=3\left(x^2-2x+1\right)-4\)
\(=3\left(x-1\right)^2-4\)
Thấy : \(3\left(x-1\right)^2\ge0\forall x\in R\)
\(\Rightarrow C=3\left(x-1\right)^2-4\ge-4\)
Vậy \(Min_C=-4\Leftrightarrow x=1\)
d, Ta có : \(D=x^2+5x+7=x^2+2.x.\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)
Thấy : \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\in R\)
\(\Rightarrow D=\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy \(Min_D=\dfrac{3}{4}\Leftrightarrow x=-\dfrac{5}{2}\)
