a)x2+2xy+y2−x−y \(\)\[= x^{2} + 2 x y + y^{2} - x - y\] \(= \left(\right. x^{2} + 2 x y + y^{2} \left.\right) - \left(\right. x + y \left.\right)\) \[=\left(\right.x+y\left.\right)^2-\left(x+y\right)\] \(=\left(\right.x+y\left.\right)\left(x+y-1\right)\)

b)2x³ + 6x² + 12x + 8

= 2(x³ + 3x² + 6x + 4)

= 2(x + 2)³

Ta có:

Suy ra A 2023 với mọi x, y

Dấu "=" xảy ra khi:

x - y + 1 = 0 y - 2 = 0x = 1 \\ y = 2

Vậy giá trị nhỏ nhất của $A$ là 2023 tại $x = 1, y = $.

a) 3�(�−1)−1+�=03x(x−1)−1+x=0

=(3x+1)(x-1)

       ⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)

       ⇒TH2:x-1=0⇒x=1

Vậy xϵ{-\(\dfrac{1}{3}\);1}

b)x²−9x=0

x(x-9)=0

         ⇒TH1:x=0

         ⇒TH2:x-9=0⇒x=9

Vậy xϵ{0;9}

a)

\(x^{2} + 25 - 10 x = x^{2} - 10 x + 25 = \left(\right. x - 5 \left.\right)^{2}\)

b)

\(- 8 y^{3} + x^{3} = x^{3} - 8 y^{3} = \left(\right. x - 2 y \left.\right) \left(\right. x^{2} + 2 x y + 4 y^{2} \left.\right)\)

a)

\(\left(\right. 2 x + 1 \left.\right)^{2} = 4 x^{2} + 4 x + 1\)
b)

\(\left(\left(\right. a - \frac{b}{2} \left.\right)\right)^{3} = a^{3} - a^{2} \frac{b}{2} + 3 a \left(\left(\right. \frac{b}{2} \left.\right)\right)^{2} - \left(\left(\right. \frac{b}{2} \left.\right)\right)^{3}\) \(= a^{3} - \frac{3}{2} a^{2} b + \frac{3}{4} a b^{2} - \frac{1}{8} b^{3}\)