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Mathematical problems of hope cup in the second day of junior high school
1 solution: from a? +b? +c? +4≤ab+3b+2c:

2(a? +b? +c? +4)-2(ab+3b+2c)=a? +b? +a? +b? +2c? +8-2ab-6b-4c

=a? -2ab+b? +a? -2a+ 1+2a-2b+(b? -4b+4)+2(c? -2c+ 1)+ 1

=(a-b)? +(a- 1)? +2(a-b)+(b-2)? +2(c- 1)? + 1

=(a-b+ 1)? +(a- 1)? +(b-2)? +2(c- 1)? ≤0

Therefore, a-b+1= a-1= b-2 = c-1= 0; Solution: a =1; b = 2; C= 1, then 200a+900b+8c=2008.

2 Solution: The point that meets the requirements is fermat point; Equation x again? The solution of-1 1x+30=0 is x1= 5; x2 = 6;

The length of three sides of isosceles △ABC is 5, 5 and 6 respectively; Or 5, 6, 6; Or 5, 5, 5; Or bhc.

(1) When the three sides of △ ABC are 5, 5 and 6 respectively, the sum of the distances from point P to vertices A, B and C is at least 4+3 √ 3;

(2) When the three sides of △ ABC are 5, 6 and 6 respectively, the sum of the distances from point P to vertices A, B and C is at least √199/2+5 √ 3/2;

(3) When the three sides of △ ABC are 5, 5 and 6 respectively, the sum of the distances from point P to the three vertices of A, B and C is at least 5 √ 3;

(4) When the three sides of △ ABC are 6, 6 and 6 respectively, the sum of the distances from point P to three vertices A, B and C is at least 6√3.

3 solution: m = 5/(a+2b); n = 7/(a-2b); Y=5/(2m)-x/2=x/2-7/(2n) Solution: x = a;; y=b

That is, the intersection point is (a, b), Mn = 35/(a? -4b? )

4 solution: 2x 4+x 3-ax? +bx+a+b- 1=2x? (x? +x-6)-x(x? +x-6)+( 13-a)x? +(b-6)x+a+b- 1

x? +x-6 is 2x 4+x 3-ax? +bx+a+b- 1, then x? +x-6 is also (13-a)x? The factor of +(b-6)x+a+b- 1

Then13-a = b-6 = (a+b-1)/(-6); Solution: a =16; b=3

5 solution: ax? +bx+c=[√ax+b/(2√a)]? +c-b? /(4a) is a completely flat path (A, B and C are constants),

What about c-b? /(4a)=0, that is, b? -4ac=0

6 solution: let x 1, x2 equation x? +MX+n = 0; Is X3 and x4 the equation y? +ny+m=0。

Then: x 1+x2 =-m, x1x2 = n; Then (x 1-x2)? =(x 1+x2)? -4x 1x2=m? -4n;

x3+x4=-n,x3x 4 = m; Then (x3-x4)? =(x3+x4)? -4x3x4=n? -4m;

By equation x? What are the two differences between +mx+n=0 and equation Y? The two differences of +ny+m=0 are equal.

Know: (x 1-x2)? =(x3-x4)? =m? -4n=n? -4m finishing: (m+n)(m-n)=-4(m-n)

And m and n are unequal real numbers, then m-n ≠ 0; So m+n=-4.

7 solution: ∵ x-2a > 0; And 6-3x ≥ 0; ∴2a<; X≤2, X has four integer solutions;

Then the four solutions are 2, 1, 0,-1; Then -2 ≤ 2a

Answer: d

note:

Fermat point's judgment

(1) For any triangle △ABC, if there is a little E in or on the triangle, if EA+EB+EC has the minimum value, then E is fermat point. Fermat point's calculation

(2) If the interior angle of a triangle is greater than or equal to 120, the vertex of this interior angle is the fermat point; If all three internal angles are less than 120, then three points with an opening angle of 120 inside the triangle are the fermat point of the triangle.