Câu hỏi của trương Đại Thành

Question

tìm min củaA=4x2-12x+11B=x2-2x+y2+4y+6tỉm max của đa thức A=-x2-6x+1B=-2x-8x-6

kudo shinichi · Accepted Answer

$A=4x^2-12x+11$$A=\left(2x\right)^2-2.2x.3+3^2+2$$A=\left(2x-3\right)^2+2$Ta có: $\left(2x-3\right)^2\ge0\forall x$$\Rightarrow\left(2x-3\right)^2+2\ge2\forall x$Dấu = xảy ra $\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}$Vậy Amin=2$\Leftrightarrow x=\frac{3}{2}$$B=x^2-2x+y^2+4y+6$$B=\left(x^2-2x+1\right)+\left(y^2+2.2y+2^2\right)+1$$B=\left(x-1\right)^2+\left(y+2\right)^2+1$Ta có:  $\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\left(y+2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\forall x;y}$Dấu = xảy ra $\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\left(y+2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-1=0\y+2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\y=-2\end{cases}}}$Vậy Bmin=1$\Leftrightarrow x=1;y=-2$$A=-x^2-6x+1$$\Rightarrow-A=x^2+6x-1$$-A=\left(x^2+2.3x+3^2\right)-10$$-A=\left(x+3\right)^2-10$$\Rightarrow A=-\left(x+3\right)^2+10$Ta có: $\left(x+3\right)^2\ge0\forall x\Rightarrow-\left(x+3\right)^2\le0\forall x\Rightarrow-\left(x+3\right)^2+10\le10\forall x$Dấu = xảy ra $\Leftrightarrow-\left(x+3\right)^2=0\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3$Vậy Amax=10$\Leftrightarrow$x= -3Sửa đề:$B=-2x^2-8x-6$$B=-2.\left(x^2+2.2x+2^2\right)+2$$B=-2.\left(x+2\right)^2+2$Ta có: $2.\left(x+2\right)^2\ge0\forall x\Rightarrow-2.\left(x+2\right)^2\le0\forall x\Rightarrow-2.\left(x+2\right)^2+2\le2\forall x$Dấu = xảy ra $\Leftrightarrow-2.\left(x+2\right)^2=0\Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2$Vậy Bmax=2$\Leftrightarrow x=-2$Câu trả lời: $A=4x^2-12x+11$$A=\left(2x\right)^2-2.2x.3+3^2+2$$A=\left(2x-3\right)^2+2$Ta có: $\left(2x-3\right)^2\ge0\forall x$$\Rightarrow\left(2x-3\right)^2+2\ge2\forall x$Dấu = xảy ra $\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}$Vậy Amin=2$\Leftrightarrow x=\frac{3}{2}$$B=x^2-2x+y^2+4y+6$$B=\left(x^2-2x+1\right)+\left(y^2+2.2y+2^2\right)+1$$B=\left(x-1\right)^2+\left(y+2\right)^2+1$Ta có:  $\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\left(y+2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\forall x;y}$Dấu = xảy ra $\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\left(y+2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-1=0\y+2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\y=-2\end{cases}}}$Vậy Bmin=1$\Leftrightarrow x=1;y=-2$$A=-x^2-6x+1$$\Rightarrow-A=x^2+6x-1$$-A=\left(x^2+2.3x+3^2\right)-10$$-A=\left(x+3\right)^2-10$$\Rightarrow A=-\left(x+3\right)^2+10$Ta có: $\left(x+3\right)^2\ge0\forall x\Rightarrow-\left(x+3\right)^2\le0\forall x\Rightarrow-\left(x+3\right)^2+10\le10\forall x$Dấu = xảy ra $\Leftrightarrow-\left(x+3\right)^2=0\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3$Vậy Amax=10$\Leftrightarrow$x= -3Sửa đề:$B=-2x^2-8x-6$$B=-2.\left(x^2+2.2x+2^2\right)+2$$B=-2.\left(x+2\right)^2+2$Ta có: $2.\left(x+2\right)^2\ge0\forall x\Rightarrow-2.\left(x+2\right)^2\le0\forall x\Rightarrow-2.\left(x+2\right)^2+2\le2\forall x$Dấu = xảy ra $\Leftrightarrow-2.\left(x+2\right)^2=0\Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2$Vậy Bmax=2$\Leftrightarrow x=-2$

ST · Answer

Đề phải là tìm min mới đúnga, A=4x2-12x+11=(4x2-12x+9)+2=(2x-3)2+2Vì (2x-3)2 $\ge$ 0 => A=(2x-3)2+2 $\ge$ 2Dấu "=" xảy ra khi 2x-3=0 <=> x=3/2Vậy Amin = 2 khi x=3/2b, B=x2-2x+y2+4y+6=(x2-2x+1)+(y2+4y+4)+1=(x-1)2+(y+2)2+1Vì $\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0$$\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0$$\Rightarrow B=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1$Dấu "=" xảy ra khi x=1,y=-2Vậy Bmin = 1 khi x=1,y=-2Câu trả lời: Đề phải là tìm min mới đúnga, A=4x2-12x+11=(4x2-12x+9)+2=(2x-3)2+2Vì (2x-3)2 $\ge$ 0 => A=(2x-3)2+2 $\ge$ 2Dấu "=" xảy ra khi 2x-3=0 <=> x=3/2Vậy Amin = 2 khi x=3/2b, B=x2-2x+y2+4y+6=(x2-2x+1)+(y2+4y+4)+1=(x-1)2+(y+2)2+1Vì $\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0$$\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0$$\Rightarrow B=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1$Dấu "=" xảy ra khi x=1,y=-2Vậy Bmin = 1 khi x=1,y=-2

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