Đáp án: $a)A_{min}=2⇔x=-7$
$b)B_{min}=-1⇔x=2$
`c)C_{min}=\frac{9}{16}⇔x=\frac{1}{2}`
$d)D_{min}=0⇔x=3$
$e)E_{min}=-16⇔x∈\{-2;1\}$
Giải thích các bước giải:
$a)$ Đặt $x+7=y$
Ta có: $A=(y+1)^4+(y-1)^4$
$=y^4+4y^3+6y^2+4y+1+y^4-4y^3+6y^2-4y+1$
$=2y^4+12y^2+2$
Do $y^4≥0∀y⇒2y^4≥0∀y$
$y^2≥0⇒12y^2≥0∀y$
$⇒2y^4+12y^2≥0∀y$
$⇒A=2y^4+12y^2+2≥2∀y$
Dấu bằng xảy ra $⇔y=0$
$⇔x+7=0⇔x=-7$
$b)B=(x-1)(x-3)(x^2-4x+5)$
$=(x^2-4x+3)(x^2-4x+5)$
$=[(x^2-4x+4)-1][(x^2-4x+4)+1]$
$=(x^2-4x+4)^2-1^2$
$=(x-2)^4-1$
Do $(x-2)^4≥0∀x$
$⇒B=(x-2)^4-1≥-1∀x$
Dấu bằng xảy ra $⇔(x-2)^4=0$
$⇔x-2=0⇔x=2$
$c)C=x^4-2x^3+3x^2-2x+1$
$=(x^4-2x^3+x^2)+(2x^2-2x)+1$
$=(x^2-x)^2+2(x^2-x)+1$
$=(x^2-x+1)^2$
`=[(x^2-x+\frac{1}{4})+\frac{3}{4}]^2`
`=[(x-\frac{1}{2})^2+\frac{3}{4}]^2`
Do `(x-\frac{1}{2})^2≥0∀x`
`⇒(x-\frac{1}{2})^2+\frac{3}{4}≥\frac{3}{4} ∀x`
`⇒C=[(x-\frac{1}{2})^2+\frac{3}{4}]^2≥(\frac{3}{4})^2=\frac{9}{16} ∀x`
Dấu bằng xảy ra `⇔(x-\frac{1}{2})^2=0`
`⇔x-\frac{1}{2}=0⇔x=\frac{1}{2}`
$d)D=x^4-6x^3+10x^2-6x+9$
$=(x^4-6x^3+9x^2)+(x^2-6x+9)$
$=x^2(x^2-6x+9)+(x^2-6x+9)$
$=(x^2+1)(x^2-6x+9)$
$=(x^2+1)(x-3)^2$
Do $(x-3)^2≥0∀x$
$x^2≥0∀x⇒x^2+1≥1>0∀x$
$⇒D=(x^2+1)(x-3)^2≥0∀x$
Dấu bằng xảy ra $⇔(x-3)^2=0$
$⇔x-3=0⇔x=3$
$e)E=(x^2+x-6)(x^2+x+2)$
$=[(x^2+x-2)-4)][(x^2+x-2)+4]$
$=(x^2+x-2)^2-16$
Do $(x^2+x-2)^2≥0∀x$
$⇒E=(x^2+x-2)^2-16≥-16∀x$
Dấu bằng xảy ra $⇔(x^2+x-2)^2=0$
$⇔x^2+x-2=0⇔(x+2)(x-1)=0$
$⇔\left[ \begin{array}{l}x+2=0\\x-1=0\end{array} \right.⇔\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.$
