$\\$
`a,`
`A=x^2 +5x+2021`
`->A=x^2 + 2 . x . 5/2 + (5/2)^2 + 8059/4`
`->A=(x+5/2)^2 + 8059/4 ≥ 8059/4∀x`
Dấu "`=`" xảy ra khi :
`(x+5/2)^2=0 ↔ x+5/2=0 ↔x=(-5)/2`
Vậy `min A=8059/4 ↔x=(-5)/2`
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`b,`
`B = (x-2)(x+3)-100`
`->B=x^2 + 3x-2x-6-100`
`->B=x^2 + x - 106`
`->B=x^2 + 2 . x . 1/2 + (1/2)^2 -425/4`
`->B=(x+1/2)^2 - 425/4 ≥ (-425)/4∀x`
Dấu "`=`" xảy ra khi :
`(x+1/2)^2= 0 ↔x+1/2=0 ↔x=(-1)/2`
Vậy `min B=(-425)/4 ↔x=(-1)/2`
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`c,`
`C=(x-1)(x+2)(x+3)(x+6)`
`-> C = [(x+2)(x+3)][(x-1)(x+6)]`
`->C= (x^2 + 3x +2x+6) (x^2 + 6x - x-6)`
`->C=(x^2 + 5x + 6)(x^2 + 5x-6)`
Đặt `x^2 + 5x +6=t`
`->C=t (t-12)`
`->C=t^2 - 12t`
`->C=t^2 - 2 . y . 6 + 6^2 - 36 = (t-6)^2 - 36 ≥ -36∀t`
Dấu "`=`" xảy ra khi :
`(t-6)^2=0 ↔t-6=0 ↔t=6`
`↔ x^2 +5x+6=6 ↔ x^2 +5x=0`
`↔ x (x+5)=0`
`↔` \(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
Vậy `min C=-36 ↔` \(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
