Tìm GTNN A=(x ²+5x+5)((x+3)(x+2)+1) B=(x-1)(x-3)(x ²-4x+5) C=2x ²+2xy+y ²-2x+10y+5 27/08/2021 Bởi Camila Tìm GTNN A=(x ²+5x+5)((x+3)(x+2)+1) B=(x-1)(x-3)(x ²-4x+5) C=2x ²+2xy+y ²-2x+10y+5
$A=(x^2+5x+5)[(x+3)(x+2)+1]$ $=(x^2+5x+5)(x^2+5x+7)$$=(x^2+5x+6-1)(x^2+5x+6+1)$$=(x^2+5x+6)^2-1≥-1$ do $(x^2+5x+6)^2≥0$ Dấu = xảy ra $⇔x^2+5x+6=0⇔x^2+2x+3x+6=0⇔(x+3)(x+2)=0$ $⇔$\(\left[ \begin{array}{l}x+3=0\\x+2=0\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}x=-3\\x=-2\end{array} \right.\) $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)$ $=(x^2-4x+4)^2-1≥-1$ do $(x^2-4x+4)^2≥0$ Dấu $=$ xảy ra $⇔x^2-4x+4⇔(x-2)^2=0⇔x-2=0⇔x=2$ $C=2x^2+2xy+y^2-2x+10y+5$ $=y^2+2y(x+5)+(x^2+10x+25)+x^2-12x+36-56$ $=y^2+2y(x+5)+(x+5)^2+(x-6)^2-56$ $=(y+x+5)^2+(x-6)^2-56≥-56$ do $(y+x+5)^2;(x-6)^2≥0$ Dấu $=$ xảy ra $⇔y+x+5=0;x-6=0⇔y+6+5=0;x=6⇔y=-11;x=6$ Bình luận
$A=(x^2+5x+5)[(x+3)(x+2)+1]$ $=(x^2+5x+5)(x^2+5x+7)$ $=(x^2+5x+6-1)(x^2+5x+6+1)$ $=(x^2+5x+6)^2-1≥-1$ do $(x^2+5x+6)^2≥0$ Dấu “=” xảy ra $⇔x^2+5x+6=0⇔x^2+2x+3x+6=0⇔(x+3)(x+2)=0$ $⇔$\(\left[ \begin{array}{l}x+3=0\\x+2=0\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}x=-3\\x=-2\end{array} \right.\) $B=(x-1)(x-3)(x^2-4x+5)$ $=(x^2-4x+3)(x^2-4x+5)$ $=(x^2-4x+4-1)(x^2-4x+4+1)$ $=(x^2-4x+4)^2-1≥-1$ do $(x^2-4x+4)^2≥0$ Dấu $=$ xảy ra $⇔x^2-4x+4$ `⇔(x-2)^2=0` `⇔x-2=0` `⇔x=2` $C=2x^2+2xy+y^2-2x+10y+5$ $=y^2+2y(x+5)+(x^2+10x+25)+x^2-12x+36-56$ $=y^2+2y(x+5)+(x+5)^2+(x-6)^2-56$ $=(y+x+5)^2+(x-6)^2-56≥-56$ do $(y+x+5)^2;(x-6)^2≥0$ Dấu $=$ xảy ra $⇔y+x+5=0;x-6=0$ `⇔y+6+5=0;x=6` `⇔y=-11;x=6` Bình luận
$A=(x^2+5x+5)[(x+3)(x+2)+1]$
$=(x^2+5x+5)(x^2+5x+7)$
$=(x^2+5x+6-1)(x^2+5x+6+1)$
$=(x^2+5x+6)^2-1≥-1$ do $(x^2+5x+6)^2≥0$
Dấu = xảy ra
$⇔x^2+5x+6=0⇔x^2+2x+3x+6=0⇔(x+3)(x+2)=0$
$⇔$\(\left[ \begin{array}{l}x+3=0\\x+2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-3\\x=-2\end{array} \right.\)
$B=(x-1)(x-3)(x^2-4x+5)$
$=(x^2-4x+3)(x^2-4x+5)$
$=(x^2-4x+4-1)(x^2-4x+4+1)$
$=(x^2-4x+4)^2-1≥-1$ do $(x^2-4x+4)^2≥0$
Dấu $=$ xảy ra $⇔x^2-4x+4⇔(x-2)^2=0⇔x-2=0⇔x=2$
$C=2x^2+2xy+y^2-2x+10y+5$
$=y^2+2y(x+5)+(x^2+10x+25)+x^2-12x+36-56$
$=y^2+2y(x+5)+(x+5)^2+(x-6)^2-56$
$=(y+x+5)^2+(x-6)^2-56≥-56$ do $(y+x+5)^2;(x-6)^2≥0$
Dấu $=$ xảy ra $⇔y+x+5=0;x-6=0⇔y+6+5=0;x=6⇔y=-11;x=6$
$A=(x^2+5x+5)[(x+3)(x+2)+1]$
$=(x^2+5x+5)(x^2+5x+7)$
$=(x^2+5x+6-1)(x^2+5x+6+1)$
$=(x^2+5x+6)^2-1≥-1$ do $(x^2+5x+6)^2≥0$
Dấu “=” xảy ra
$⇔x^2+5x+6=0⇔x^2+2x+3x+6=0⇔(x+3)(x+2)=0$
$⇔$\(\left[ \begin{array}{l}x+3=0\\x+2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-3\\x=-2\end{array} \right.\)
$B=(x-1)(x-3)(x^2-4x+5)$
$=(x^2-4x+3)(x^2-4x+5)$
$=(x^2-4x+4-1)(x^2-4x+4+1)$
$=(x^2-4x+4)^2-1≥-1$ do $(x^2-4x+4)^2≥0$
Dấu $=$ xảy ra $⇔x^2-4x+4$
`⇔(x-2)^2=0`
`⇔x-2=0`
`⇔x=2`
$C=2x^2+2xy+y^2-2x+10y+5$
$=y^2+2y(x+5)+(x^2+10x+25)+x^2-12x+36-56$
$=y^2+2y(x+5)+(x+5)^2+(x-6)^2-56$
$=(y+x+5)^2+(x-6)^2-56≥-56$ do $(y+x+5)^2;(x-6)^2≥0$
Dấu $=$ xảy ra $⇔y+x+5=0;x-6=0$
`⇔y+6+5=0;x=6`
`⇔y=-11;x=6`