\(A=x^2-4x+1\)
\(A=x^2-4x+4-3\)
\(A=\left(x-2\right)^2-3\)
Min A = -3
Min A xảy ra khi (x-2)²=0
                           x-2=0
                           x=2
 

A đến C là tìm GTNN

\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)

Dấu "=" xảy ra ⇔ x=2

\(B=2x^2-x+1=2\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)+\dfrac{7}{8}=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{4}\)

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

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

D đến F là tìm GTLN

\(E=-x^2+2x-2=-\left(x^2-2x+1\right)-1=-\left(x-1\right)^2-1\le-1\)

Do (x-1)²≥0 ⇔-(x-1)²≤0 ⇔ D≤-1

Dấu "=" xảy ra ⇔ x=1

\(D=-x^2+x-3=-\left(x^2-2.\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{11}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

\(F=-3x^2+x-2=-3\left(x^2-2.\dfrac{1}{6}+\dfrac{1}{36}\right)-\dfrac{23}{12}=-3\left(x-\dfrac{1}{6}\right)-\dfrac{23}{12}\le-\dfrac{23}{12}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{6}\)

\(A=x^2-4x+1=\left(x^2-4x+4\right)-3=\left(x-2\right)^2-3\ge-3\)

Vậy \(A_{Min}=-3khix=2\)

Bài 5:

a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=2\)

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

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Bài 4:

a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxM=7\Leftrightarrow x=2\)

b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(A=x^2+x\) . Có: \(x^2\ge x\Rightarrow x^2+x\ge0\)

Dấu '=' xảy ra khi: \(x^2+x=0\Rightarrow x\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy: \(Min_A=0\) tại \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

\(B=4x-12x+10\)

\(B=-8x+10\)

\(B=10-8x\)

Xét: \(x< 0\Rightarrow10-8x\ge10\)

Dấu '=' xảy ra khi: \(8x=0\Rightarrow x=0\)

Xét: \(x>0\Rightarrow10-8x\le10\)

Dấu '=' xảy ra khi: \(8x=0\Rightarrow x=0\)

Vậy: Khi x<0. \(Min_B=10\) tại \(x=0\)

Khi: x>0.  \(Max_B=10\)tại \(x=0\)

K chắc

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

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

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

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

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

\(A=x^2+12x+36=x^2+12x+36+3=\left(x+6\right)^2+3\ge3\)

Dấu '=' xảy ra khi x=-6

\(B=9x^2-12x+4-4=\left(3x-2\right)^2-4\ge-4\)

Dấu '=' xảy ra khi x=2/3

\(C=-x^2+4x+1\)

\(=-\left(x^2-4x-1\right)=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2