a)
\(A=3x^2+15x=3(x^2+5x)=3(x^2+5x+\frac{5^2}{2^2})-\frac{75}{4}\)
\(=3(x+\frac{5}{2})^2-\frac{75}{4}\)
Vì \((x+\frac{5}{2})^2\geq 0, \forall x\Rightarrow A\geq 3.0-\frac{75}{4}=-\frac{75}{4}\)
Vậy \(A_{\min}=-\frac{75}{4}\Leftrightarrow x=-\frac{5}{2}\)
c)
\(C=x^2-2x+y^2-4y+7\)
\(=(x^2-2x+1)+(y^2-4y+4)+2\)
\(=(x-1)^2+(y-2)^2+2\)
Vì \((x-1)^2\geq 0; (y-2)^2\geq 0, \forall x,y\Rightarrow C\geq 0+0+2=2\)
Vậy \(C_{\min}=2\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\end{matrix}\right.\)
b)
\(B=(x-1)(x+2)(x+3)(x+6)\)
\(=[(x-1)(x+6)][(x+2)(x+3)]\)
\(=(x^2+5x-6)(x^2+5x+6)\)
\(=a(a+12)\) (đặt \(x^2+5x-6=a\) )
\(=(a^2+12a+36)-36=(a+6)^2-36\)
Vì \((a+6)^2\geq 0, \forall a\Rightarrow B\geq 0-36=-36\)
Vậy \(B_{\min}=-36\). Dấu "=" xảy ra khi \(a+6=0\Leftrightarrow x^2+5x-6+6=0\)
\(\Leftrightarrow x^2+5x=0\Leftrightarrow x=0\) hoặc $x=-5$