It is known that two α, β of the quadratic equation anx2 − an + 1x + 1 = 0 (n ∈ n *) of X satisfy 6 α - 2 α β + 6 β = 3, and A1 = 1. (1) use an to express an + 1; (2) find the general term formula of sequence an; (3) find the first n term and Sn of sequence {an} and the value range of Sn

It is known that two α, β of the quadratic equation anx2 − an + 1x + 1 = 0 (n ∈ n *) of X satisfy 6 α - 2 α β + 6 β = 3, and A1 = 1. (1) use an to express an + 1; (2) find the general term formula of sequence an; (3) find the first n term and Sn of sequence {an} and the value range of Sn

(1) ∵ two α, β, ∵ α + β = an + 1An, α β = 1An, ∵ 6 α - 2 α β + 6 β = 3, ∵ 6 · an + 1an-2 · 1An = 3 ∵ an + 1 = 12An + 13 of the quadratic equation anx2-an + 1x + 1 = 0 (n ∈ n *); (2) it is known from (1) that an + 1-23 = 12 (an-23), ∵ {an-23} takes 13 as the leading term

Quadratic equation
It is known that a > 0 equation AX2 (this 2 represents the second power of x) + 4bx + 4C = 0 has two real solutions on [2,3]
1. Prove the existence of a triangle with A.B.C as its edge
2. Prove a / (a + C) + B / (B + a) > C / (B + C)
Wrong ax2-4bx + 4C = O

1. Certification:
Because the equation has two solutions on [2,3], X1 + x2 = 4B / a > = (2 + 2) = 4, X1 * x2 = 4C / a > = 2 * 2 = 4
Because a > 0, b > = a, C > = a
Let's prove that there is a triangle with A.B.C as the side. From the above analysis, we only need to prove that there is a triangle with A.B.C as the side
A + b > C, a + C > B. that is 4 + (x1 + x2) > X1 * X2, and 4 + X1 * x2 > X1 + x2
4> | X1 + x2-x1 * x2 | let's prove that there are many ways to prove this formula
G (x1) = X1 + x2-x1 * X2, so I just need to prove that the minimum value of G (x1) is greater than - 4 and the maximum value is less than 4. Since g (x1) is a linear function of x1, I just need to compare the function values when X1 = 2 and X1 = 3
G (2) = 2-x2, G (3) = 3-2 * X2, we can see that no matter what the value of X2 is, the absolute value of G (x1) is less than 4, so we finally get the conclusion!
2 prove: from a + b > C, if BC / (a + C)
From C > b > = a, we can see that B / (a + C) C / (B + C) is obtained by substituting
If b > = C, then divide both sides by (a + b), then a / (a + C) > = A / (a + b), C / (a + b) > = C / (B + C)
Also available
So, no matter what, the original inequality holds!
If you don't understand, send me a message

Can mathematical quadratic equation a [500 - (A-25)] = 13500 be solved?
It's a [500-10 (A-25)] = 13500

a^2-525a+13500=0
Δ=525^2-4×13500=262125
a=(525±512)/2
a1=519.5,a2=6.5

Natural number n plus 2 is a complete square number, and natural number n minus 1 is also a complete square number

Let a ^ 2 = n + 2; B ^ 2 = n-1
Then a ^ 2-B ^ 2 = 3 → (a + b) (a-b) = 3 = 3 × 1
If a and B are positive integers, then a + B = 3, A-B = 1
The simultaneous equations, a = 2, B = 1
So n = 2

Please explain the reasons for listing the following equations,
The resistance of 10 ohm resistance wire will increase in different degrees in a and B environments. One section of resistance wire in a environment and four sections of resistance wire in B environment are connected in parallel, and the resistance is 33 / 14. Two sections of resistance wire in a environment and three sections of resistance wire in B environment are connected in parallel, and the resistance is 44 / 19

1 / x + 4 / y = 14 / 33, 2 / x + 3 / y = 19 / 44, y = 12, x = 11, the resistance increases by 1 in a environment and 2 in B environment

Application of binary quadratic equations
A school bought a total of 25 sets of books with the same price in the bookstore and Dangdang. Because online shopping can enjoy a certain discount, the price of each set on Dangdang is 10 yuan cheaper. It is known that Dangdang spent 1350 yuan on books, 350 yuan more than the bookstore. How many sets of books did the school buy in the bookstore and Dangdang?

Suppose the school bought X and Y sets of books in the bookstore and Dangdang. Com
x+y=25
1350/x +10 = （1350-350）/y
x = 15
y= 10

How to solve the quadratic equation of two variables
I haven't read for a long time. I can't read at all
3+m=3p
3 + 5m = 3P ~ (2)
The answer is m = 0, P = 1 and M = 9, P = 4
How do you calculate this?

By substituting the first Formula M = 3p-3 into the second formula, we can get 3 + 5 (3p-3) = 3P ^ 2, then we can get P = 1 or P = 4. The corresponding M = 0 or M = 9

Solve the quadratic equation of two variables!
X+X^=13
Find x =?
X+X^2=13
Find x =?

First of all, it's not a bivariate quadratic equation, it's a bivariate quadratic equation
Then, the equation can be rewritten as the square of (x + 1 / 2) equals 53 / 4
That is, x + 1 / 2 = plus or minus 53 / 4
Namely
X = 51 / 4 or - 55 / 4

How to solve bivariate quadratic equation
For example, 1. XY + X + y = - 13, x ^ 2 + y ^ 2 = 29
2.x^2-3xy-4y^ 2=0,x@2+4xy+4y ^2=1
Two questions should be x ^ 2-3xy-4y ^ 2 = 0, x ^ 2 + 4xy + 4Y ^ 2 = 1

xy+x+y=-13,x^2+y^2=29→（x+y）^2-2xy=29
By eliminating xy and regarding x + y as a whole, we can reduce the degree
x^2-3xy-4y^2=0,…… （1）
x^2+4xy+4y^2=1…… (2)
From (1), we can get (x-4y) (x + y) = 0,
That is, x = 4Y or x = - Y
From (2), we can get (x + 2Y) ^ 2 = 1, that is, x + 2Y = 1 or x + 2Y = - 1
The four binary linear equations can be formed into equations respectively

A method for solving quadratic equations of square two variables
For example, if the system of equations x + y = 6, xy = 8, then we can get Z & sup2; - (x + y) Z + xy = 0 and then borrow Z1 = 2, Z2 = 4, then X1 = 2, Y1 = 4, X2 = 4, Y1 = 2. What is the name of this formula? Is there any other similar method for solving special binary quadratic equation?

Weida theorem
Other methods:
Substituting x = 6-y into xy = 8 leads to
y(6-y)=8
The solution is y ^ 2-6y + 8 = 0, Y 1 = 2, y 2 = 4
So X1 = 4, X2 = 2