Given A2 + A + 1 = 0, find the value of A3 + 2A2 + 2A + 1,

Given A2 + A + 1 = 0, find the value of A3 + 2A2 + 2A + 1,

a3+2a2+2a+1
=a^3+a^2+a+a^2+a+1
=a(a^2+a+1)+(a^2+a+1)
=(a^2+a+1)(a+1)
=0

Simplification of (2A / A2-4) x ((A2 + 4 / 4A) - 1)

2a/(a^2-4)x[(a^2+4)/(4a)-1]
=2a/(a^2-4)x[(a^2+4-4a)/(4a)]
=2a/[(a-2)(a+2)]x(a-2)^2/(4a)
=2/(a+2)x(a-2)/4
=2(a-2)/[4(a+2)]
=(a-2)/[2(a+2)]
=(a-2)/(2a+4)

[a ^ 2 + 4A] ^ 2 + 8 [a ^ 2 + 4A] + 16 factorization process
The factorization I want is the process

Let x = A & sup2; + 4a
Then the original formula = x & sup2; + 8x + 16
=(x+4)²
=(a²+4a+4)²
=[(a+2)²]²
=(a+2)^4

How to factorize 16 + 8 (a ^ + 4a) + (a ^ + 4a) ^?

16+8(a^+4a)+(a^+4a)^
=4^+4*2*(a^+4a)+(a^+4a)^
=[4+(a^+4a)]^
=[(a+2)^]^
=(a+2)(a+2)(a+2)(a+2)

If the sum of the two parts of the equation x2-ax-2a = 0 is 4a-3, then the product of the two parts is______ ．

From the meaning of the question, we know that X1 + x2 = a = 4a-3, ∧ a = 1, ∧ x1x2 = - 2A = - 2, so fill in - 2

If the sum of the two squares of the equation x-ax-4a = 0 about X is the square-3 of 4A, find the product of the two equations

x²-ax-4a=0
Then X1 + x2 = a
So a = 4A & # 178; - 3
4a²-a-3=0
(4a+3)(a-1)=0
a=-3/4,a=1
a=-3/4
x²+3/4x+4=0
△

If the sum of the two equations of x ^ 2-ax-4a = 0 is 4A ^ 2-3, find the product of the two equations

If there are two equations, then
If △ = a ^ 2 + 16A ≥ 0, the constant holds
According to Veda's theorem,
a=4a^2-3
4a^2-a-3=0
(a-1)(4a+3)=0
A = 1 or a = - 3 / 4
The product of two is - 4A
When a = 1, the product of two is - 4
When a = - 3 / 4, the product of two is 3

How to solve the equation 4A + 4B = 32 - B = 48

4a+4b=32
a+b=8
a²－b²＝48
(a+b)(a-b)=48
a-b=6
∴a=7,b=1

4a2 (the square of 4A) = a, how to solve this equation

4a²=a
4a²-a=0
a(4a-1)=0
a1=0,a2=¼

Let f (x) = (1 + a) x ^ 4 + x ^ 3 - (3a + 2) x ^ 2-4a. Let f (x) = (1 + a) x ^ 4 + x ^ 3 - (3a + 2) x ^ 2-4a

f(x)
=(1+a)x^4+x^3-(3a+2)x^2-4a
=(x^4+x^3-2x^2)+(ax^4-3ax^2-4a)
=(x^2+x-2)x^2+a(x^4-3x^2-4)
=(x+2)(x-1)x^2+a(x^2-4)(x^2+1)
=(x+2)(x-1)x^2+a(x+2)(x-2)(x^2+1)
=(x+2)[(x-1)x^2+a(x-2)(x^2+1)]
For any real number a, there exists x0, constant f (x0) ≠ 0, that is, for any real number a, there exists x0, constant f (x0) has nothing to do with the value of a, its f (x0) ≠ 0
Obviously, there are x + 2 = 0 and (X-2) (x ^ 2 + 1) = 0, and f (x) has nothing to do with a value, but f (x) ≠ 0, so only (X-2) (x ^ 2 + 1) = 0 is consistent
When (X-2) (x ^ 2 + 1) = 0, f (x) has nothing to do with a, then x = 2