Câu hỏi của Huyền Anh

Question

TÌm GTNN:1) 2x2 + 9y2 - 6xy - 6x - 12y + 2004.2) x( x + 1)( x2 + x - 4).3) ( x2 + 5x + 5)[( x + 2)( x + 3) + 1].4) ( x - 1)(x - 3)( x2 - 4x + 5)HELP ME !!!!!

Lightning Farron · Accepted Answer

1)$2x^2+9y^2-6xy-6x-12y+2004$$=x^2+x^2-6xy+9y^2-6x-12y+2004$$=x^2+\left(x-3y\right)^2-10x+4x-12y+2004$$=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+2004$$=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+4+25+1975$$=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x^2-10x+25\right)+1975$$=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975$Dấu "=" khi $\begin{cases}\left(x-5\right)^2=0\\left(x-3y+2\right)^2=0\end{cases}$$\Leftrightarrow\begin{cases}x=5\y=\frac{7}{3}\end{cases}$Vậy Min=1975 khi $\begin{cases}x=5\y=\frac{7}{3}\end{cases}$2)$x\left(x+1\right)\left(x^2+x-4\right)=\left(x^2+x\right)\left(x^2+x-4\right)$Đặt $t=x^2+x$ ta có:$t\left(t-4\right)=t^2-4t+4-4$$=\left(t-2\right)^2-4\ge-4$Dấu "=" khi $t-2=0\Leftrightarrow t=2\Leftrightarrow x^2+x=2$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\x=1\end{array}\right.$Vậy Min=-4 khi $\left[\begin{array}{nghiempt}x=-2\x=1\end{array}\right.$3)$\left(x^2+5x+5\right)\left[\left(x+2\right)\left(x+3\right)+1\right]$$=\left(x^2+5x+5\right)\left[x^2+5x+6+1\right]$Đặt $t=x^2+5x+5$ ta có:$t\left(t+1\right)=t^2+t+\frac{1}{4}-\frac{1}{4}=\left(t+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}$Dấu "=" khi $t+\frac{1}{2}=0\Leftrightarrow t=-\frac{1}{2}\Leftrightarrow x^2+5x+5=-\frac{1}{2}$$\Leftrightarrow x_{1,2}=\frac{-10\pm\sqrt{12}}{4}$Vậy Min=$-\frac{1}{4}$ khi $x_{1,2}=\frac{-10\pm\sqrt{12}}{4}$4)$\left(x-1\right)\left(x-3\right)\left(x^2-4x+5\right)$$=\left(x^2-4x+3\right)\left(x^2-4x+5\right)$Đặt $t=x^2-4x+3$ ta có:$t\left(t+2\right)=t^2+2t+1-1=\left(t+1\right)^2-1\ge-1$Dấu "=" khi $t+1=0\Leftrightarrow t=-1\Leftrightarrow x^2-4x+3=-1\Leftrightarrow x=2$Vậy Min=-1 khi x=2   Câu trả lời: 1)$2x^2+9y^2-6xy-6x-12y+2004$$=x^2+x^2-6xy+9y^2-6x-12y+2004$$=x^2+\left(x-3y\right)^2-10x+4x-12y+2004$$=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+2004$$=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+4+25+1975$$=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x^2-10x+25\right)+1975$$=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975$Dấu "=" khi $\begin{cases}\left(x-5\right)^2=0\\left(x-3y+2\right)^2=0\end{cases}$$\Leftrightarrow\begin{cases}x=5\y=\frac{7}{3}\end{cases}$Vậy Min=1975 khi $\begin{cases}x=5\y=\frac{7}{3}\end{cases}$2)$x\left(x+1\right)\left(x^2+x-4\right)=\left(x^2+x\right)\left(x^2+x-4\right)$Đặt $t=x^2+x$ ta có:$t\left(t-4\right)=t^2-4t+4-4$$=\left(t-2\right)^2-4\ge-4$Dấu "=" khi $t-2=0\Leftrightarrow t=2\Leftrightarrow x^2+x=2$$\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\x=1\end{array}\right.$Vậy Min=-4 khi $\left[\begin{array}{nghiempt}x=-2\x=1\end{array}\right.$3)$\left(x^2+5x+5\right)\left[\left(x+2\right)\left(x+3\right)+1\right]$$=\left(x^2+5x+5\right)\left[x^2+5x+6+1\right]$Đặt $t=x^2+5x+5$ ta có:$t\left(t+1\right)=t^2+t+\frac{1}{4}-\frac{1}{4}=\left(t+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}$Dấu "=" khi $t+\frac{1}{2}=0\Leftrightarrow t=-\frac{1}{2}\Leftrightarrow x^2+5x+5=-\frac{1}{2}$$\Leftrightarrow x_{1,2}=\frac{-10\pm\sqrt{12}}{4}$Vậy Min=$-\frac{1}{4}$ khi $x_{1,2}=\frac{-10\pm\sqrt{12}}{4}$4)$\left(x-1\right)\left(x-3\right)\left(x^2-4x+5\right)$$=\left(x^2-4x+3\right)\left(x^2-4x+5\right)$Đặt $t=x^2-4x+3$ ta có:$t\left(t+2\right)=t^2+2t+1-1=\left(t+1\right)^2-1\ge-1$Dấu "=" khi $t+1=0\Leftrightarrow t=-1\Leftrightarrow x^2-4x+3=-1\Leftrightarrow x=2$Vậy Min=-1 khi x=2

Huyền Anh · Answer

Thank you !Câu trả lời: Thank you !

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