The positions of real numbers a, B and C on the number axis are shown in the figure. Try to simplify | a | - | a + B | + | C-A | + | b-c| .b.a.0.c.

The positions of real numbers a, B and C on the number axis are shown in the figure. Try to simplify | a | - | a + B | + | C-A | + | b-c| .b.a.0.c.

|a|-|a+b|+|c-a|+|b-c|
=-a+a+b+c-a+c-b
=-a+2c

How to multiply (3x + 2) by (3x-2)?
Number, algorithm

(3x + 2) times (3x-2) = the square of 3x-2 = 9x * x-4

A. It is proved that the eigenvalues of AB are all greater than zero

First of all, Pt represents the transpose of P matrix, and P-1 represents the inverse of P matrix
In this paper, the sufficient and necessary condition for a real symmetric matrix A to be a positive definite matrix is that there exists an invertible matrix P such that
A = PTP "
If a and B are positive definite, then there exists an invertible matrix P, Q such that
A = PTP
B = QTQ
Q(AB)Q-1 = Q(PTP)(QTQ)Q-1=QPTPQT = (PQT)T(PQT)
P. Q is reversible, so pqt is also an invertible matrix,
Again, by using the necessary and sufficient conditions at the beginning, q (AB) Q-1 is a positive definite matrix and all eigenvalues are greater than zero
Because q is an invertible matrix, AB is similar to the matrix Q (AB) Q-1, so the eigenvalues of AB are all greater than zero
OK, that's it,

The limit of sequence {an} is A. It is proved that the limit of (a1 + A2 +... + an) / N = a

LIM (n - > ∞) an = a, prove: LIM (n - > ∞) (a1 + A2 +.. + an) / N = a
prove:
① For any ε > 0,
∵ lim(n->∞) an =a
For ε / 2 > 0, there exists N1. When n > N1, | an-a | Max {m, N1}:
|(a1+a2+..+an)/n - a|
≤ (|a1-a|+|a2-a|+...+|aN1-a|)/n +(|a(N1+1)-a|+...+|an-a|)/n
≤ ε/2 +(n-N1)*ε/2/n ≤ ε/2+ε/2 = ε
② So there exists n = max {[M], N1} ∈ Z+
③ When n > N,
④ It is found that | (a 1 + a 2 +.. + an) / N - a | < e holds
∴ lim(n->∞) (a1+a2+..+an)/n=a
The simplest way to solve this problem is to apply o'stoltz theorem directly
The converse proposition does not hold
an = (-1)^n
LIM (n - > ∞) (a1 + A2 +.. + an) / N = 0, but:
An = (- 1) ^ n divergence

4x+6x=10

Solution 10x = 10
X=10÷10
x=1

It is known that the definition field of function f (x) is {x | x ≠ 0, X ∈ r}. For any X1 and X2 in the definition field, f (x 1 times x 2) = f (x 1) + F (x 2), and if x > 1, f (x) is greater than 0
(1) Finding the value of F (1) and f (- 1)
(2) To prove that f (x) is an even function
(3) To prove that f (x) is an increasing function on (0, positive infinity)

(1)
Substituting X1 = 1, X2 = 1 into f (x1 * x2) = f (x1) + F (x2)
So f (1) = 0
Substituting X1 = - 1, X2 = - 1 into f (x1 * x2) = f (x1) + F (x2)
We get f (- 1) = 0
(2)
Substituting x2 = - 1 into f (x1 * x2) = f (x1) + F (x2)
Then f (- x1) = f (x1)
F (x) is an even function
(3)
Let x1 ∈ (0, + ∞), X2 > 1, then (x1 * x2) ∈ (0, + ∞) and X1 * x2 > x1
F (x1 * x2) = f (x1) + F (x2)
That is, f (x1 * x2) - f (x1) = f (x2)
∵x2>1
∴f(x2)>0
That is, f (x1 * x2) - f (x1) > 0
Ψ f (x) is an increasing function on (0, + ∞)

Find the set of all zeros of function f (x) = 2x Cube - 3x + 1

2x³-3x+1
=2x³-2x-x+1
=2x(x+1)(x-1)-(x-1)
=(x-1)(2x²+2x-1)=0
x=1,
2x+2x-1=0
x=(-1±√3)/2
So it's {1, (- 1 + √ 3) / 2, (- 1 - √ 3) / 2}

If the perimeter of the rectangle is 4 root sign 3 and the length of its diagonal is 2 root sign 2, then the area of the rectangle is_________

Let one side of a long shape be x, then the other side be 2 √ 3-x,
According to Pythagorean theorem
X^2+(2√3-X)^2=(2√2)^2,
X^2-2√3X+2=0,
According to Weida's theorem, we can get: S = 2

If x + y = 4, X & # 178; + Y & # 178; = 14, find the value of XY

Solution
x+y=4
The square of both sides is as follows:
x²+2xy+y²=16
∵x²+y²=14
∴2xy+14=16
∴2xy=2
∴xy=1

Solving equation 7 + (x-1) * 1 = 2x / 2 + 1.5x/2

Solving the equation
7+(X-1)*1=2X/2+1.5X/2
The results are as follows
7+x-1=3.5x/2
14+x-3.5x-2=0
-2.5x=-12
x=4.8