Is the absolute value of √ 3 - √ 5 irrational?

Is the absolute value of √ 3 - √ 5 irrational?

The absolute value of √ 3 - √ 5 is irrational

Please explain rational numbers and irrational numbers in combination with high school practice!
The meaning of real number set!

If a number can be transformed into a fraction, and the numerator and denominator are integers, then the number is rational, such as 1.25 = 5 / 4, 4 = 4 / 1, 0.01 = 1 / 100, etc. otherwise, it is irrational, such as root 2, π, etc. these irrational numbers can be proved. Rational numbers and irrational numbers are called real numbers

If you add an integer to itself, you want to subtract it, multiply it, divide it, and add up the sum, difference, product, and quotient to 64, what is the number

Let this number be X
2x+0+x^2+1=64
x^2+2x-63=0
(x+9)(x-7)=0
X = - 9 or x = 7

As shown in the figure, it is known that the side length of a large square is a + B + C. using the area relation of the figure, we can get: (a + B + C) 2 = A2 + B2 + C2 + 2Ab + 2BC + 2Ac. When the side length of a large square is a + B + C + D, we can get: (a + B + C + D) 2 = A2 + B2 + C2 + D2 + 2Ab + 2Ac + 2ad + 2BC + 2bd + 2CD by using the area relation of the figure. Generally, the square of the sum of N numbers is equal to the sum of the squares of these n numbers plus two times of their products According to the above conclusions, the following problems can be solved: (1) if a + B + C = 6, A2 + B2 + C2 = 14, then AB + BC + AC=______ (2) multiply any two of the five numbers - 4, - 2, - 1, 3, 5, and then add all the products. If the sum is m, find the value of M

(1) If a + B + C = 6 is the square of both sides, then, (a + B + C) 2 = A2 + B2 + C2 + 2Ab + 2BC + 2Ac = 36, ｜ AB + BC + AC = [36 - (A2 + B2 + C2)] 2 = (36-14) △ 2 = 11; (2) ∵ - 4-2-1 + 3 + 5 = 1, ｜ after the square of both sides, (- 4-2-1 + 3 + 5) 2 = 42 + 22 + 12 + 32 + 52 + 2m = 55 + 2m = 1, ｜ M = (1-55) △ 2 = - 54 △ 2 = - 27

The sum of three A's is 3a, and the product of three A's multiplication is A3______ ．

The sum of three A's is 3a, and the product of three A's multiplication is A3

At least 25 mathematical problems about cross multiplication
The harder it is, the better
The main problem is calculation, and the auxiliary problem is application

Let 2x & sup2; - 7x + 3 decompose the factor 5x2 + 6xy-8y2
X-Y) (2x-2y-3) - 2 factorization

There are four numbers in the equal ratio sequence. The multiplication of four numbers equals 16. The sum of the two numbers in the middle is 5?

Let the first be a and the common ratio be Q
The multiplication of four numbers is a ^ 4 * q ^ 6 = 16, and the middle two numbers are only AQ + AQ ^ 2 = 5
The first root a ^ 2 * q ^ 3 = 4 = > AQ * AQ ^ 2 = 4
We immediately get that one is 1 and the other is 4, so the common ratio is 4, and the first number is 1 / 4
So four numbers are 1 / 4 1 4 16

Math problem (cross multiplication)
1. The square of X - 250x - 375000 = 0
2. The square of Y - 34y + 240 = 0

1.(x-750)*(x+500)=0
x=750
x=-500
2.(y-24)*(y-10)=0
y=24
y=10

(m+n)(m-n)(m^2+n^2)
I learned how to multiply polynomials on the first day of junior high school

(m+n)(m-n)(m^2+n^2)
=（m^2-n^2)(m^2+n^2)
=m^4-n^4

(m-n)^2 +n(n- m)
(a-b)(3r-t)+9(b-a)(2t-5s)

(m-n)^2 +n(n- m)
Original formula = (n-m) ^ 2 + n (n-m)
=(n- m)（n-m+n）
=(n-m)(2n-m)
Because of the complete square, so (m-n) ^ 2 = (n-m) ^ 2
(a-b)(3r-t)+9(b-a)(2t-5s)
The original formula = (a-b) (3r-t) - 9 (a-b) (2t-5s)
=(a-b)（3r-t-18t+45s）
=(a-b)（3r-19t+45s）