Tìm min và max cả 2 câu hả bạn

\(A=x^2-6x+11=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2 \)

Vậy GTNN của A là 2 khi x = 3

\(B=2x^2+10x-1=2\left(x^2+5x+\frac{25}{4}\right)-\frac{27}{2}=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge-\frac{27}{2}\)

Vậy GTNN của B là \(-\frac{27}{2}\)khi x = \(-\frac{5}{2}\) 

a, \(2x^2-4xy+4y^2-6x\)

\(=x^2-2xy-2xy+4y^2+x^2-3x-3x+9-9\)

\(=\left(x-2y\right)^2+\left(x-3\right)^2-9\)

Với mọi giá trị của \(x;y\in R\) ta có:

\(\left(x-2y\right)^2+\left(x-3\right)^2-9\ge-9\)

Để \(\left(x-2y\right)^2+\left(x-3\right)^2-9=-9\) thì

\(\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3-2y=0\\x=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1,5\\x=3\end{matrix}\right.\)

Vậy..............

b, \(z^2-4zt+5t^2-2t+13\)

\(=z^2-2zt-2zt+4t^2+t^2-t-t+1+12\)

\(=\left(z-2t\right)^2+\left(t-1\right)^2+12\)

Với mọi giá trị của \(z;t\in R\) ta có:

\(\left(z-2t\right)^2+\left(t-1\right)^2+12\ge12\)

Để \(\left(z-2t\right)^2+\left(t-1\right)^2+12=12\) thì

\(\left\{{}\begin{matrix}\left(z-2t\right)^2=0\\\left(t-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z-2=0\\t=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=2\\t=1\end{matrix}\right.\)

Vậy...............

Câu c tường tự !!!

a,Đặt A= \(2x^2-4xy+4y^2-6x\)

\(=\left(2x^2-4xy-6x\right)+4y^2\)

\(=2\left(x^2-2xy-3x\right)+4y^2\)

\(=2\left[x^2-2x\left(y+\dfrac{3}{2}\right)+\left(y+\dfrac{3}{2}\right)^2\right]+4y^2-\left(y+\dfrac{3}{2}\right)^2\)

\(=2\left(x-y-\dfrac{3}{2}\right)^2+4y^2-y^2-3y-\dfrac{9}{4}\)

\(=2\left(x-y-\dfrac{3}{2}\right)^2+3\left(y^2-y+\dfrac{1}{4}\right)-3\)

\(=2\left(x-y-\dfrac{3}{2}\right)^2+3\left(y-\dfrac{1}{2}\right)^2-3\)

Với mọi giá trị của x;y ta có:

\(\left(x-y-\dfrac{3}{2}\right)^2\ge0;\left(y-\dfrac{1}{2}\right)^2\ge0\)

\(\Rightarrow2\left(x-y-\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2-3\ge-3\)

Vậy Min A = -3 khi \(\left\{{}\begin{matrix}x-y-\dfrac{3}{2}=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}-\dfrac{3}{2}=0\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

b, Đặt B = \(z^2-4zt+5t^2-2t+13\)

\(=\left(z^2-4zt+4t^2\right)+\left(t^2-2t+1\right)+12\)

\(=\left(z-2t\right)^2+\left(t-1\right)^2+12\)

Với mọi giá trị của z;t ta có:

\(\left(z-2t\right)^2\ge0;\left(t-1\right)^2\ge0\)

\(\Rightarrow\left(z-2t\right)^2+\left(t-1\right)^2+12\ge12\)

Vậy Min B = 12 khi \(\left\{{}\begin{matrix}z-2t=0\\t-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z-2=0\\t=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=2\\t=1\end{matrix}\right.\)

c, Đặt C = \(16x^2-8x+y^2-2y\)

\(=\left(16x^2-8x+1\right)+\left(y^2-2y+1\right)-2\)

\(=\left(4x-1\right)^2+\left(y-1\right)^2-2\)

Với mọi giá trị x;y ta có:

\(\left(4x-1\right)^2\ge0;\left(y-1\right)^2\ge0\)

\(\Rightarrow\left(4x-1\right)^2+\left(y-1\right)^2-2\ge-2\)

Vậy Min C = -2 khi \(\left\{{}\begin{matrix}4x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=1\end{matrix}\right.\)

\(x^2-6x+15\)

\(=x^2-6x+9+6\)

\(=\left(x-3\right)^2+6\)

\(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-3\right)^2+6\ge6\)

Dấu "=" xảy ra khi:
\(\left(x-3\right)^2=0\Rightarrow x=3\)

\(x^2-6x+15=x^2+2.x.3+3^2+6=\left(x+3\right)^2+6\le6\)

Vậy: Min x² - 6x + 15 = 6

Check kq bằng máy tính fx 570 VN PLUS:

Ấn theo thứ tự:

Mode tới 5 tới 3 ấn 1 = -6 = 15 = = = = KQ = 6.

ta có : \(x^2-6x+15=x^2-2.x.3+3^2+6=\left(x+3\right)^2+6\ge6\) với mọi x

\(\Rightarrow Min\) của \(x^2-6x+15\) là 6 khi \(\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

vậy GTNN của \(x^2-6x+15\) là 6 khi \(x=-3\)

\(A=x^2-6x+15\)

\(A=x^2-2\cdot x\cdot3+3^2+6\)( biến đổi về dạng HĐT )

\(A=\left(x-3\right)^2+6\)

vì ( x - 3 )² luôn >= 0 với mọi x

\(\Rightarrow A\ge6\)với mọi x

Dấu "=" xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy Amin = 6 <=> x = 3

\(B=2x^2-10x+8\)

\(B=2\left(x^2-5x+4\right)\)

\(B=2\left(x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{9}{4}\right)\)

\(B=2\left[\left(x-\frac{5}{2}\right)^2-\frac{9}{4}\right]\)

\(B=2\left(x-\frac{5}{2}\right)^2-\frac{9}{2}\)

Vì 2( x - 5/2 )² luôn >= 0 với mọi x

\(\Rightarrow B\ge\frac{-9}{2}\)với mọi x

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy Bmin = -9/2 <=> x = 5/2

a, \(m^2-6m+x^2-x+3\)

\(=m^2-3m-3m+9+x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{25}{4}\)

\(=\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\)

Với mọi giá trị của \(m;x\in R\) ta có:

\(\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\)

Để \(\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}=-\dfrac{25}{4}\) thì

\(\left\{{}\begin{matrix}\left(m-3\right)^2=0\\\left(x-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy..............

b, \(3x^2-6x+12\)

\(=3x^2-3x-3x+3+9\)

\(=3x\left(x-1\right)-3\left(x-1\right)+9\)

\(=3\left(x-1\right)^2+9\)

Với mọi giá trị của \(x\in R\) ta có:

\(3\left(x-1\right)^2+9\ge9\)

Để \(3\left(x-1\right)^2+9=9\) thì

\(\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy..............

Chúc bạn học tốt!!!

a, \(A=m^2-6m+x^2-x+3\)

\(=x^2-6m+9+x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{25}{4}\)

\(=\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\ge\dfrac{-25}{4}\)

Dấu " = " khi \(\left\{{}\begin{matrix}\left(m-3\right)^2=0\\\left(x-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(MIN_A=\dfrac{-25}{4}\) khi m = 3, \(x=\dfrac{1}{2}\)

b, \(B=3x^2-6x+12=3\left(x^2-2x+4\right)\)

\(=3\left(x^2-2x+1+3\right)=3\left(x-1\right)^2+9\ge9\)

Dấu " = " khi \(3\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy MIN B = 9 khi x = 1