a: \(A=2x^2-8x+1\)

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

\(=2\left(x^2-4x+4-\dfrac{7}{2}\right)\)

\(=2\left(x-2\right)^2-7>=-7\)

Dấu = xảy ra khi x=2

b: \(B=\left(x-3\right)^2+\left(x-1\right)^2\)

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

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

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

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

Dấu = xảy ra khi x=2

a,\(x^2-6x-17=x^2-2\cdot3x+9-26=\left(x-3\right)^2-26\ge-26\)

b, \(x^2-10x=x^2-2\cdot5x+25-25=\left(x-5\right)^2-25\ge-25\)

c,\(3x^2-12x+5=3x^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+12-7=\left(\sqrt{3}x-2\sqrt{3}\right)^2-7\ge-7\)

d,\(2x^2-x-1=2x^2-2\cdot\sqrt{2}x\cdot\dfrac{1}{2\sqrt{2}}+\dfrac{1}{8}-\dfrac{9}{8}=\left(\sqrt{2}x-\dfrac{1}{2\sqrt{2}}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

e,\(x^2+y^2-8x+4y+27=x^2-2\cdot4x+16+y^2+2\cdot2y+4+7=\left(x-4\right)^2+\left(y+2\right)^2+7\ge7\)

f,\(x\left(x-6\right)=x^2-6x=x^2-2\cdot3x+9-9=\left(x-3\right)^2-9\ge-9\)

h,\(\left(x-2\right)\cdot\left(x-5\right)\cdot\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x-10\right)=\left(x^2-7x\right)^2-100\ge-100\)

Mình giúp tính biểu thức thôi

còn lại bạn tự làm nhé

Bài 4.

a) 3xy² - 45x²y = 3xy( y - 15x )

b) 25y² - 4x² + 4x - 1

= 25y² - ( 4x² - 4x + 1 )

= ( 5y )² - ( 2x - 1 )²

= ( 5y - 2x + 1 )( 5y + 2x - 1 )

c) x² - 5x + xy - 5y

= x( x - 5 ) + y( x - 5 )

= ( x - 5 )( x + y )

d) x² - 8x - 33

= x² + 3x - 11x - 33

= x( x + 3 ) - 11( x + 3 )

= ( x + 3 )( x - 11 )

Bài 5.

a) A = ( x - 2 )³ - x²( x - 4 ) + 8

= x³ - 6x² + 12x - 8 - x³ + 4x² + 8

= -2x² + 12x

B = ( x² - 6x + 9 ) : ( x - 3 ) - x( x + 7 ) - 9

= ( x - 3 )² : ( x - 3 ) - x² - 7x - 9

= x - 3 - x² - 7x - 9

= -x² - 6x - 12

b) Với x = -1 thì A = -2.(-1)² + 12.(-1) = -2 - 12 = -14

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 

Giups mik vs

1. Ta có \(|3x-1|=\frac{1}{2}\)

\(\Rightarrow\)\(\orbr{\begin{cases}3x-1=\frac{1}{2}\\3x-1=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=(\frac{1}{2}+1):3\\x=(-\frac{1}{2}+1):3\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)

Sau đó tự thay x vào đa thức theo 2 trường hợp trên nha

Sai thì thôi nha bn mik cx chưa lm dạng này bh

Câu 1:

\(A\left(x\right)=6x^4-4x^2-3+9x+5x^2-7x-2x^4+4-2x-4x^4\)

\(=\left(6x^4-2x^4-4x^4\right)+\left(-4x^2+5x^2\right)+\left(-7x-2x\right)+9x+\left(-3+4\right)\)

\(=x^2+9x+1\)

Ta có: \(\left|3x-1\right|=\frac{1}{2}\)

TH1: \(3x-1=\frac{1}{2}\Rightarrow3x=\frac{1}{2}+1=\frac{3}{2}\Rightarrow x=\frac{3}{2}:3=\frac{1}{2}\)

\(A\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+9\cdot\frac{1}{2}+1=\frac{1}{4}+\frac{9}{2}+1=\frac{23}{4}\)

TH2: \(3x-1=\frac{-1}{2}\Rightarrow3x=\frac{-1}{2}+1=\frac{1}{2}\Rightarrow x=\frac{1}{2}:3=\frac{1}{6}\)

\(A\left(\frac{1}{6}\right)=\left(\frac{1}{6}\right)^2+9\cdot\frac{1}{6}+1=\frac{91}{36}\)

Câu 2:

Theo đề, ta có:

\(B\left(x\right)=\left(2x+3\right)^2+1=3\frac{7}{9}\)

\(\left(2x+3\right)^2=\frac{34}{9}-1=\frac{25}{9}\)

TH1: \(2x+3=\frac{-5}{3}\Rightarrow2x=\frac{-5}{3}-3=-\frac{14}{3}\)

TH2: \(2x+3=\frac{5}{3}\Rightarrow2x=\frac{5}{3}-3=\frac{-4}{3}\)