a, Ta có: x2 ≥ 0 => x2 + 3x ≥ 0
=> x2 + 3x + 19 ≥ 19
Dấu "=" xảy ra <=> x2 + 3x = 0
<=> x(x + 3) = 0
<=> x = 0 hoặc x = -3
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a, \(A=x^2+3x+19\)
\(A=x^2+\frac{3}{2}\cdot2x+\frac{9}{4}+\frac{67}{4}\)
\(A=\left(x+\frac{3}{2}\right)^2+\frac{67}{4}\)
\(\left(x+\frac{3}{2}\right)^2\ge0\Rightarrow\left(x+\frac{3}{2}\right)^2+\frac{67}{4}\ge\frac{67}{4}\)
\(\Rightarrow A\ge\frac{67}{4}\)
dấu "=" xảy ra khi :
\(\left(x+\frac{3}{2}\right)^2=0\Rightarrow x+\frac{3}{2}=0\Rightarrow x=-\frac{3}{2}\)
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