8

2³.6 - 72 : 3²

= 8.6 - 72 : 9

= 48 - 8

= 40

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a chia 7 dư 3

=>a-3⋮7

=>a-3+56⋮7

=>a+53⋮7(1)

a chia 9 dư 1

=>a-1⋮9

=>a-1+54⋮9

=>a+53⋮9(2)

Từ (1),(2) suy ra a+53∈BC(7;9)

=>a+53∈B(63)

=>a+53∈{63;126;189;252;315;378;441;504;...}

=>a∈{10;73;136;199;262;325;388;451;...}

mà 350<a<500

nên a∈{388;451}

mà a⋮11

nên a=451

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

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

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