It is known that two real roots of the equation (A2-1) X2 - (a + 1) x + 1 = 0 about X are reciprocal to each other, then the value of a is___ ．

It is known that two real roots of the equation (A2-1) X2 - (a + 1) x + 1 = 0 about X are reciprocal to each other, then the value of a is___ ．

∵ the equation (A2-1) X2 - (a + 1) x + 1 = 0 has two real roots, ∵ a ≠± 1. Let the two real roots of equation (A2-1) X2 - (a + 1) x + 1 = 0 be α and β respectively, and ∵ the two real roots of equation (A2-1) X2 - (a + 1) x + 1 = 0 are reciprocal to each other, ∵ α β = 1a2-1 = 1. The solution is a = ± 2, ∵ △ = [- (a + 1

M quartic - 2m square + 1

The fourth power of M - 2m & # 178; + 1
=(m²-1)²
=[(m+1)(m-1)]²
=(m+1)²(m-1)²

Given a nonempty set a = {X / X * 2-4mx + 2m + 6 = 0., X ∈ r} if a is included in the range of positive real number M

∵ the set a = {x | X & sup2; - 4mx + 2m + 6 = 0} is not an empty set, the equation x & sup2; - 4mx + 2m + 6 = 0 has a solution, that is (- 4m) & sup2; - 4 (2m + 6) ≥ 0, m ≤ - 1, or m ≥ 3 / 2, ∵ the set a = {x | X & sup2; - 4mx + 2m + 6 = 0} is contained in the positive real number set, the roots of the equation x & sup2; - 4mx + 2m + 6 = 0 are positive numbers, ∵ 2m > 0

A mathematical problem to be solved by binary linear equations
There is only one car for a class to go for an outing in Beishan Mountain, which is 18 kilometers away. It needs to be divided into two groups. Group A takes the bus first, group B walks. When the car arrives at a, group A gets off and walks, and the car returns to meet group B. the last two groups arrive at Beishan Mountain at the same time. The known speed of the car is 60 km / h, and the walking speed is 4 km / h

Suppose the distance from point a to Beishan is x, that is, the walking distance of group A. because the two groups arrive at the same time, the walking distance of the two groups is the same, and the driving distance is the same. The distance from point a to meeting group B is 18-2x

Can a & sup2; + B & sup2; + C & sup2; + 2ab-2ac + 2BC decompose a factor?

No

Decomposition factor: ① A & sup2; + B & sup2; + C & sup2; + 2ab-2ac-2bc; ② a-sixth power + b-sixth power

Factorization A & sup2; - 3B & sup2; - 8C & sup2; + 2Ab + 2Ac + 14bc

（a-3b+4c)(a-3b-2c)
Here!

Factorization of B & # 178; + C & # 178; + 2Ab + 2Ac + 2BC
2. Factorization of x-178; + X - (a-178; - a)

b²+c²+2ab+2ac+2bc
=b²+2bc +c²+2ab+2ac
=(b+c)²+2a(b+c)
=(b+c)(b+c+2a)
x²+x-（a²-a）
=x²+x-a²+a
=x²-a²+x+a
=(x+a)(x-a)+(x+a)
=(x+a)(x-a+1)

(- 150A ^ 6B ^ 2C) / (- 3A ^ 4B ^ 2) calculation

=50a²c

Given the absolute value of (3x-4b-11) + (7a + 6B + 5) square = 0, then a =, B=

The absolute value of (3a-4b-11) + the square of (7a + 6B + 5) = 0
So 3a-4b-11 = 0 (1)
7a+6b+5=0 (2)
(1)×3+(2)×2
9x-33+14a+10=0
23a=23
therefore
a=1
b=(3a-11)/4=-2