`A=6x -3x^2`
`A=-3x^2+6x-3+3`
`A=-(3x^2-6x+3)+3`
`A=-3(x^2-2x+1)+3`
`A=-3(x-1)^2+3`
Có: `(x-1)^2\ge0⇒-3(x-1)^2\le0⇒-3(x-1)^2+3\le3`
Dấu ''='' xảy ra khi: `x-1=0⇔x=1.`
Vậy `Max_A=3⇔x=1.`
`B=4y -y^2`
`B=-y^2+4y-4+4`
`B=-(y^2-4y+4)+4`
`B=-(y-2)^2+4`
Có: `(y-2)^2\ge0⇒-(y-2)^2\le0⇒-(y-2)^2+4\le4`
Dấu ''='' xảy ra khi: `y-2=0⇔x=2.`
Vậy `Max_B=4⇔x=2.`
`C=(x-1)(x-2)(x-3)(x-4) +2`
`C=[(x-1)(x-4)][(x-2)(x-3)]+2`
`C=(x^2-5x+4)(x^2-5x+6)+2`
Đặt `x^2-5x+5=y`
`⇒C=(y+1)(y-1)+2`
`C=y^2-1+2`
`C=y^2+1`
Có: `y^2\ge0⇒y^2+1\ge1`
Dấu ''='' xảy ra khi: `y=0⇔x^2-5x+5=0`
`⇔4(x^2-5x+5)=0.4`
`⇔4x^2-20x+20=0`
`⇔(4x^2-20x+25)-5=0`
`⇔ (2x-5)^2 - (\sqrt{5})^2=0`
`⇔ (2x-5-\sqrt{5})(2x-5+\sqrt{5})=0`
`⇒` \(\left[ \begin{array}{l}2x-5-\sqrt{5}=0\\2x-5+\sqrt{5}=0\end{array} \right.\)
`⇒ x= \frac{5±\sqrt{5}}{2}`
Vậy $Min_C$`=1⇔x= \frac{5±\sqrt{5}}{2}.`
`D=(x+1)(x+2)(x+5)(x+6)-3`
`D=[(x+1)(x+6)][(x+2)(x+5)]-3`
`D=(x^2+7x+6)(x^2+7x+10)-3`
Đặt `x^2+7x+8=y`
`⇒ D=(y+2)(y-2) - 3`
`D=y^2-4-3`
`D=y^2-7`
Có: `y^2\ge0⇒y^2-7\ge-7`
Dấu ''='' xảy ra khi: `y=0⇔x^2+7x+8=0`
`⇔4(x^2+7x+8)=4.0`
`⇔4x^2+28x+32=0`
`⇔(4x^2+28x+49)-17=0`
`⇔(2x+7)^2-(\sqrt{17})^2=0`
`⇔(2x+7-\sqrt{17})(2x+7+\sqrt{17})=0`
`⇒` \(\left[ \begin{array}{l}2x+7-\sqrt{17}=0\\2x+7+\sqrt{17}=0\end{array} \right.\)
`⇒ x=\frac{-7±\sqrt{17}}{2}`
Vậy $Min_D$`=-7⇔x= \frac{-7±\sqrt{17}}{2}.`
