a) Đặt \(A=x^2+6x=x^2+6x+9-9=\left(x+3\right)^2-9\)
Vì \(\left(x+3\right)^2\ge0\forall x\Rightarrow\left(x+3\right)^2-9\ge-9\)
''='' xảy ra khi \(x+3=0\Rightarrow x=-3\)
Vậy \(A_{MIN}=-9\) khi x = -3
b) Đặt \(B=x^2+3x-5=x^2+2\cdot x\cdot1,5+2,25-\dfrac{29}{4}\)
\(=\left(x+1,5\right)^2-\dfrac{29}{4}\)
Vì \(\left(x+1,5\right)^2\ge0\forall x\Rightarrow\left(x+1,5\right)^2-\dfrac{29}{4}\ge-\dfrac{29}{4}\)
''='' xảy ra khi x + 1,5 = 0 => x = -1,5
Vậy \(B_{MIN}=-\dfrac{29}{4}\) khi \(x=-1,5\)
c) Đặt \(C=x^2+17x+6=x^2+2\cdot x\cdot8,5+72,25-\dfrac{265}{4}\)
\(=\left(x+8,5\right)^2-\dfrac{265}{4}\)
Vì \(\left(x+8,5\right)^2\ge0\forall x\Rightarrow\left(x+8,5\right)^2-\dfrac{265}{4}\ge-\dfrac{265}{4}\)
''='' xảy ra khi x = -8,5
Vậy=...
