a)

A = -4x² - 12x

= -4(x² + 3x)

b)

B = 3 - 4x - x²

= -(x² + 4x - 3)

= -(x² + 4x + 4 - 7)

= -(x + 2)² + 7

Do (x + 2)² ≥ 0

⇒ -(x + 2)² ≤ 0

⇒ -(x + 2)² + 7 ≤ 7

Vậy maxB = 7 khi x = -2

\(x^{64}+x^{32}+1\\ =x^{64}+2x^{32}+1+x^{32}-2x^{32}\\ =\left[\left(x^{32}\right)^2+2\cdot x^{32}\cdot1+1^2\right]-x^{32}\\ =\left(x^{32}+1\right)^2-\left(x^{16}\right)^2\\ =\left(x^{32}-x^{16}+1\right)\left(x^{32}+x^{16}+1\right)\\ =\left(x^{32}-x^{16}+1\right)\left[\left(x^{32}+2x^{16}+1\right)+x^{16}-2x^{16}\right]\\ =\left(x^{32}-x^{16}+1\right)\left[\left(x^{16}+1\right)^2-x^{16}\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^{16}+x^8+1\right)\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left[\left(x^{16}+2x^8+1\right)-x^8\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left[\left(x^8+1\right)^2-x^8\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^8+x^4+1\right)\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left[\left(x^8+2x^4+1\right)-x^4\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left[\left(x^4+1\right)^2-x^4\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\)

\(=\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left[\left(x^4+2x^2+1\right)-x^2\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left[\left(x^2+1\right)^2-x^2\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)

\(x^{64}+x^{32}+1\)

\(=x^{64}+2x^{32}-x^{32}+1\)

\(=\left(x^{64}+2^{32}+1\right)-x^{32}\)

\(=\left(x^{32}+1\right)^2-\left(x^{16}\right)^2\)

\(=\left(x^{32}+1-x^{16}\right)\left(x^{32}+1+x^{16}\right)\)

\(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=\left(x^2-7+3\right)\left(x^2-7-3\right)-72\)

\(=\left(x^2-7\right)^2-9-72\)

\(=\left(x^2-7\right)^2-81\)

\(=\left(x^2-7+9\right)\left(x^2-7-9\right)\)

\(=\left(x^2+2\right)\left(x^2-16\right)\)

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

\(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-10x^2-4x^2+40-72\)

\(=x^4-14x^2-32\)

\(=\left(x^2-16\right)\left(x^2+2\right)=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

\(x^4+6x^3+7x^2-6x+1\\ =\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)

`= x^2(x^2 + x + 1) - x(x^2 + x + 1) + 2024(x^2 + x + 1)`

`= (x^2 - x + 2024)(x^2 + x + 1)`.

\(3\cdot\left(2\cdot1\right)=6x-3x\\ 6=3x\\ x=\dfrac{6}{3}=2\)

a: \(101^2=\left(100+1\right)^2=100^2+2\cdot100\cdot1+1^2\)

=10000+200+1

=10201

b: \(64^2+36^2+72\cdot64\)

\(=64^2+2\cdot64\cdot36+36^2\)

\(=\left(64+36\right)^2=100^2=10000\)

c: \(54^2+46^2-2\cdot54\cdot46=\left(54-46\right)^2=8^2=64\)

d: \(98\cdot102=\left(100-2\right)\left(100+2\right)=100^2-4=9996\)

cuu

Yêu cầu của đề bài là gì vậy bạn?

`a) 2024x - 2024y - 2024x^2 + 4048xy - 2024y^2`

`= ( 2024x - 2024y) - (2024x^2 - 4048xy + 2024y^2)`

`= 2024 (x-y) - 2024 (x^2 - 2xy + y^2)`

`= 2024 (x-y) - 2024 (x-y)^2`

`= 2024 (x-y) (1 - x + y)`

`b) x^2 + 5x - 6`

`= (x^2 - x) + (6x - 6)`

`= x(x-1) + 6(x-1)`

`= (x+6)(x-1)`

`c) 2x^2 + 3x - 5`

`= (2x^2 - 2x) + (5x - 5)`

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

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

`d) x^4 + 4 `

`= (x^2)^2 + 2^2`

`= (x^2)^2 + 4x^2 + 2^2 - 4x^2`

`= (x^2 + 2)^2 - (2x)^2 `

`= (x^2 - 2x + 2)(x^2 + 2x+ 2)`

`e) x^5 + x + 1`

`= (x^5 - x^2) + (x^2 + x + 1)`

`= x^2 (x^3 - 1) + (x^2 + x + 1)`

`= x^2 (x-1) (x^2 + x + 1) + (x^2 + x + 1)`

`= (x^3 - x^2 + 1) (x^2 + x + 1) `