bài 1:= \(2x\left(x-3\right)-6\left(x-3\right)+2y\left(x-3\right)\)

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

bài 2:P=\(x^2-2x+1+y^2+6y+9+2\)

         P=\(\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

vậy Pmin=2 khi x=1 và y=-3

ở oooo

e: \(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

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

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

c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)

d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

a) Ta có: \(x^2-2x-4y^2-4y\)

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

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

b) Ta có: \(x^3+2x^2+2x+1\)

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

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

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

d) Ta có: \(a^6-a^4+2a^3+2a^2\)

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

\(=a^2\left[a^2\left(a^2-1\right)+\left(2a+2\right)\right]\)

\(=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]\)

\(=a^2\cdot\left(a+1\right)\left(a^3-a+2\right)\)

c) Ta có: \(x^3-4x^2+12x-27\)

\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)

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

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

Cách làm: trước hết rút gọn đa thức, sau đó tìm hạng tử có bậc cao nhất ⇒ bậc của đa thức.

⇒ Bậc 2 là bậc cao nhất.

⇒ Đa thức có bậc 2 (bậc của đa thức là bậc của hạng tử có bậc cao nhất).

a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)

b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)

c: \(2x-1-x^2\)

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

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

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

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

\(=\left(5-x\right)\left(5+3x\right)\)

h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)

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

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

i: \(=x^2y^2-4xy+4-3\)

\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)

k: \(=y^2-\left(x-1\right)^2\)

\(=\left(y-x+1\right)\left(y+x-1\right)\)

l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)

m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)

a) Ta có: \(25x^2-20x+7\)

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

\(=\left(5x-2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{2}{5}\)

b) Ta có: \(9x^2-6x+2\)

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

\(=\left(3x-1\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

c) Ta có: \(-x^2+2x-2\)

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

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

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

Dấu '=' xảy ra khi x-1=0

hay x=1

d) Ta có: \(x^2+12x+39\)

\(=x^2+12x+36+3\)

\(=\left(x+6\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=-6

e) Ta có: \(-x^2-12x\)

\(=-\left(x^2+12x+36-36\right)\)

\(=-\left(x+6\right)^2+36\le36\forall x\)

Dấu '=' xảy ra khi x=-6

f) Ta có: \(4x-x^2+1\)

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

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

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

Dấu '=' xảy ra khi x=2

a) Ta có: \(25x^2-20x+7\)

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

\(=\left(5x-2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{2}{5}\)

b) Ta có: \(9x^2-6x+2\)

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

\(=\left(3x-1\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

c) Ta có: \(-x^2+2x-2\)

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

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

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

Dấu '=' xảy ra khi x=1

( Mình trình bày mẫu câu a các câu khác mình làm tắt lại nhưng tương tự trình bày câu a nha )

a, Ta có : \(25x^2-20x+7=\left(5x\right)^2-2.5x.2+2^2+3\)

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

Thấy : \(\left(5x-2\right)^2\ge0\forall x\in R\)

\(\Rightarrow\left(5x-2\right)^2+3\ge3\forall x\in R\)

Vậy \(Min=3\Leftrightarrow5x-2=0\Leftrightarrow x=\dfrac{2}{5}\)

b, \(=9x^2-2.3x+1+1=\left(3x-1\right)^2+1\ge1\)

Vậy Min = 1 <=> x = 1/3

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

Vậy Max = -1 <=> x = 1

d, \(=x^2+2.x.6+36+3=\left(x+6\right)^2+3\ge3\)

Vậy Min = 3 <=> x = - 6

e, \(=-x^2-2.x.6-36+36=-\left(x+6\right)^2+36\le36\)

Vậy Max = 36 <=> x = -6 .

f, \(=-x^2+4x-4+5=-\left(x^2-4x+4\right)+5=-\left(x-2\right)^2+5\le5\)

Vậy Max = 5 <=> x = 2

2.

a. 3x(12x - 4) - 9x(4x - 3) = 30

<=> 36x² - 12x - 36x² + 27x = 30

<=> 36x² - 36x² - 12x + 27x = 30

<=> 15x = 30

<=> x = 2

b. x(5 - 2x) + 2x(x - 1) = 15

<=> 5x - 2x² + 2x² - 2x = 15

<=> -2x² + 2x² + 5x - 2x = 15

<=> 3x = 15

<=> x = 5

a) x² ( 5x³ - x - 2323x²y=  6969x³y²- 2323x⁴y+ 2323x²y²

c) x² ( 4x^{3 -} 5xy + 2x ) ( -

b: \(=\left(x-y\right)^2-4y^2\)

\(=\left(x-y-2y\right)\left(x-y+2y\right)\)

\(=\left(x-3y\right)\left(x+y\right)\)

c: \(=x\left(x-6\right)+y\left(x-6\right)\)

\(=\left(x-6\right)\left(x+y\right)\)