`b)`
`A=x^2+5x+10`
`A=x^2+2 . x. 5/2+25/4-25/4+10`
`A=(x^2+2. x. 5/2+25/4)-25/4+10`
`A=(x+5/2)^2+15/4`
Ta có:`(x+5/2)^2≥0∀x`
`⇒(x+5/2)^2+15/4≥15/4∀x`
Vậy `A_{min}=15/4` khi `x+5/2=0⇔x=-5/2`
`d)`
`C=(x-1)(x+2)(x+3)(x+6)`
`C=[(x-1)(x+6)][(x+2)(x+3)]`
`C=(x²+6x-x-6)(x²+3x+2x+6)`
`C=(x^2+5x-6)(x^2+5x+6)`
`C=[(x^2+5x)-6][(x^2+5x)+6]`
`C=(x^2+5x)^2-6^2`
`C=(x^2+5x)^2-36`
Ta có:`(x^2+5x)^2≥0∀x`
`⇒(x^2+5x)^2-36≥-36∀x`
Vậy `C_{min}=-36` khi `x^2+5x=0⇔x(x+5)=0⇔`\(\left[ \begin{array}{l}x=0\\x+5=0\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
