Ta có: A = x² - 5x + 1 = (x² - 5x + 25/4) - 21/4 = (x - 5/2)² - 21/4

Ta luôn có: (x - 5/2)² \(\ge\)0 \(\forall\)x

=> (x - 5/2)² - 21/4 \(\ge\)-21/4 \(\forall\)x

Dấu "=" xảy ra <=> x -5/2 = 0 <=> x = 5/2

Vậy Min A = -21/4 tại  x = 5/2

Ta có: B = -x + 3x + 1 = -(x - 3x  + 9/4) + 13/4 = -(x - 3/2)² + 13/4

Ta luôn có: -(x - 3/2)² \(\le\)0 \(\forall\)x

=> -(x - 3/2)² + 13/4 \(\le\)13/4 \(\forall\)x

Dấu "=" xảy ra <=> x - 3/2 = 0 <=> x  = 3/2

Vậy Max B = 13/4 tại x = 3/2

(xem lại đề)

b)  (2x-6)(x+4)=0

c)  (x-3)(x+4)<0

d)  (x+2)(X-5)>0

bạn đăg tách ra cho m.n cùng giúp nhé

Bài 2 : 

a, \(A=\left|2x-4\right|+2\ge2\)

Dấu ''='' xảy ra khi x = 2 

Vậy GTNN A là 2 khi x = 2 

b, \(B=\left|x+2\right|-3\ge-3\)

Dấu ''='' xảy ra khi x = -2 

Vậy GTNN B là -3 khi x = -2 

\(C=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)

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

\(\ge x-1+0+3-x=2\)

Dấu "=" khi \(\begin{cases}x-1\ge0\\x-2=0\\x-3\le0\end{cases}\)\(\Rightarrow\begin{cases}x\ge1\\x=2\\x\le3\end{cases}\)\(\Rightarrow x=2\)

Vậy \(Min_C=2\) khi x=2

d)\(D=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\)

\(=\left|x-1\right|+\left|x-2\right|+\left|3-x\right|+\left|4-x\right|\)

\(\ge x-1+x-2+3-x+4-x=4\)

Dấu "=" khi \(\begin{cases}x-1\ge0\\x-2\ge0\\3-x\ge0\\4-x\ge0\end{cases}\)\(\Rightarrow\begin{cases}x\ge1\\x\ge2\\x\le3\\x\le4\end{cases}\)\(\Rightarrow2\le x\le3\)

Vậy \(Min_D=4\) khi \(2\le x\le3\)

\(A=x^2-6x+11\)

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

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

\(\Rightarrow A\ge2\)

\(\Rightarrow MinA=2\)

\(Khi\)\(\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\)

Chúc bn học giỏi nhoa!!!

ban kia lam dung roi do

k tui nha

thanks

Bài 2: 

a: Ta có: \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

b: Ta có: \(-x^2+x+2\)

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

\(=-\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}\right)\)

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

f: Ta có: \(x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi x=1 và y=2

e: Ta có: \(3x^2-6x+1\)

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

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

\(=3\left(x-1\right)^2-2\ge-2\forall x\)

Dấu '=' xảy ra khi x=1

Bài 1: 

a: Ta có: \(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2\cdot\left[\left(x+3\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)

b: Ta có: \(x^3-3x+2=0\)

\(\Leftrightarrow x^3-x-2x+2=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

\(A=2006-\frac{x}{6-x}\le2006\)

Min \(A=2006\Leftrightarrow\frac{x}{6-x}=0\Rightarrow x=0\)

\(B=\left|x-2001\right|+\left|x+1\right|\ge0\)

Min \(B=0\Leftrightarrow\hept{\begin{cases}x-2001=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2001\\x=-1\end{cases}}}\)