Note: √ is the root sign, after X and Y? 、? Is it after the code, number, a, (n+ 1)? It's square

1、

① Test center: Solving quadratic equation of one variable-formula method.

② Analysis: Firstly, the root of the equation is expressed according to the formula method, and then the hypothesis method is used to analyze that A is a positive integer and T is a rational number, so T is an integer.

3 solution: solution: equation about x x? The root of -2ax+b=0 is a √ a? -b, the equation about y? The root of +2ay+b=0 is -a √ a? -B.

Set √a? -b=t, then

When x? =a+t，x？ = a-t； y？ =-a+t，y？ =-a-t, and x? y？ -x？ y？ =0, the condition is not met;

When x? =a-t，x？ = a+t； y？ =-a-t，y？ =-a+t, and x? y？ -x？ y？ =0, the condition is not met;

When x? =a-t，x？ = a+t； y？ =-a+t，y？ =-a-t，x？ y？ -x？ y？ = 4at

When x? =a+t，x？ = a-t； y？ =-a-t，y？ =-a+t，x？ y？ -x？ y？ =-4at。

Because t=√a? -b > 0, so there is at=502. ..

(10)

Because a is a positive integer, we know that t is a rational number, so t is an integer.

From at=502, a=25 1, t=2, that is, the minimum value of b is b=a? -t？ =25 1? -2? =62997.

So the minimum value of b is 62997.

(15)

④ Comments: This question mainly examines the formula solution of quadratic equation with one variable, which is quite difficult. After finding the root, it is the key to solve the problem to analyze the qualified values of B respectively.

2、

① Test center: inscribed circle and center of triangle; The area of a triangle.

② Analysis: Let ∠A=2∠B, there should be a? =b(b+c), and a > B. when a > c > b, let a=n+ 1, c=n, b=n- 1, and substitute a? =b(b+c)，get (n+ 1)？ =(n- 1)？ (2n- 1), the length of three sides can be obtained.

3 solution: solution: let ∠A=2∠B, there should be a? =b(b+c), and a > B. When a > c > b, let a=n+ 1, c=n, b=n- 1, (n is a positive integer greater than 1).

Substitute a? =b(b+c)，get

(n+ 1)？ =(n- 1)？ (2n- 1)，

The solution is n=5,

∴a=6,b=4,c=5.

Comments: This is a comprehensive question, which is difficult to examine the inscribed circle and area of a triangle.