Đáp án:
Giải thích các bước giải:
a) `A=x^2+3x+7`
`A=x^2 + 2. 3/2 x+9/4+19/4`
`A=(x+3/2)^2+19/4`
Ta có: `(x+3/2)^2 \ge 0 \forall x`
`⇒ (x+3/2)^2+19/4 \ge 19/4 \forall x`
Vậy `A_{min}=19/4` khi `x+3/2=0⇔x=-3/2`
b) `B=4x^2-4x-5`
`B=4x^2-4x+1-6`
`B=(2x-1)^2-6`
Ta có: `(2x-1)^2 \ge 0 \forall x`
`⇒ (2x-1)^2-6 \ge -6 \forall x`
Vậy `B_{min}=-6` khi `2x-1=0⇔x=1/2`
c) `C=(x-2)(x-5)(x^2-7x-10)`
`C=(x^2-5x-2x+10)(x^2-7x-10)`
`C=(x^2-7x+10)(x^2-7x-10)`
Đặt `x^2-7x=t`
`C=(t+10)(t-10)`
`C=t^2-100`
Ta có: `t^2 \ge 0 \forall t`
`⇒ t^2-100 \ge -100 \forall t`
Vậy `C_{min}=-100` khi `t=0`
`⇔ x^2-7x=0`
`⇔ x(x-7)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x=7\end{array} \right.\)
