It is known that a, B, C are the three sides of the triangle ABC, and it is proved that: (a? + B? - C?)? - 4A? B

It is known that a, B, C are the three sides of the triangle ABC, and it is proved that: (a? + B? - C?)? - 4A? B

0

0

(a? + B? - C?)? - 4A? B? = (a? + B? + 2ab-c?) = [(a + b)? C?] [(a-b) 2? - C?] = (a + B + C) (a-b + C) (a-b-c) because, a, B and C are the three

It is known that a, B and C are the three sides of △ ABC respectively, and it is proved that: (a? + B? - C?) 2-4a? B? 2 < 0

decompose
（a2+b2-C2+2ab）（a2+b2-c2-2ab）=（a+b-c）（a+b+c）（a-b-c）（a-b+c）
The sum of any two sides of the triangle is greater than the third day
It is known that only a-b-c is less than 0, others are greater than 0, so the syndrome is less than 0

Given that a, B, C are the three side lengths of △ ABC, please determine the positive and negative values of the algebraic formula (a? + B? - C?) - 4A? B

The original formula = (a 2 + B 2 - C 2 + 2 ab) (a 2 + B 2 - C 2 - 2 ab)
=[(a+b)²-c²][(a-b)²-c²]
=(a+b+c)(a+b-c)(a-b+c)(a-b-c)
Obviously, a + B + C > 0
The sum of the two sides of the triangle is greater than the third
So a + B-C > 0
a-b+c>0
a-b-c

If a, B, C are the three side lengths of △ ABC, then is the value of the algebraic formula (a? B? - C?) - 4A? B? Be positive or negative

The first three are positive

It is known that the three sides of an unequal triangle ABC are integers ABC, and satisfy the requirements that the square of a + the square of D - 4a-6b + 13 = 0, find the length! High reward of C

A 2 - 4A + B 2 - 6B + 14 = a? - 4A + 4 + B? - 6B + 9 = (A-2) 2 + (B-3) 2 = 0, a = 2 b = 3
Because the three sides in the condition are all integers and are not equal to each other, the difference between the two sides of the triangle is greater than that of the third side, and the difference between the two sides is less than that of the third side. Ha ha ha ha

It is known that the three side lengths a, B and C of △ ABC are integers, and a and B satisfy the requirements A − 2 + B 2 − 6B + 9 = 0

A kind of
a−2+b2-6b+9=
a−2+（b-3）2=0，
∴a-2=0，b-3=0，
That is, a=2, b=3,
﹤ 3-2 ﹤ C ﹤ 3 + 2, i.e. 1 ﹤ C ﹤ 5,
Then C = 2,3,4

Given that a, B, C are integers and A2 + B2 + C2 + 48 < 4A + 6B + 12C, then (1 A+1 B+1 c) The value of ABC is______ .

∵ a, B, C are integers,
∴a2+b2+c2+48≥48，
Both sides of the original inequality are positive integers,
⇔ the inequality A2 + B2 + C2 + 48 < 4A + 6B + 12C ⇔ A2 + B2 + C2 + 48 + 1 ≤ 4A + 6B + 12C,
∴（a-2）2+（b-3）2+（c-6）2≤0，
Qi
a−2＝0
b−3＝0
c−6＝0 ，
The solution is,
a＝2
b＝3
c＝6 ，
∴(1
A+1
B+1
c)abc=1；
So the answer is: 1

If we know that the length of the three sides of the unequal edge △ ABC is a positive integer a, B, C, and satisfies A2 + b2-4a-6b + 13 = 0, then the length of C edge is () A. 2 B. 3 C. 4 D. 5

∵a2+b2-4a-6b+13，
=a2-4a+4+b2-6b+9，
=（a-2）2+（b-3）2=0，
∴a-2=0，b-3=0，
A = 2, B = 3,
∵3-2=1，3+2=5，
∴1＜c＜5，
The length of the three sides of ABC is positive integers a, B, C,
∴c=4．
Therefore, C

It is known that the lengths of the three sides of the unequal triangle ABC are integers a, B, C, and satisfy the value of C of a ^ 2 + B ^ 2-4a-6b + 13 = 0

According to the equation (A-2) + (B-3) = 0, a = 2, B = 3. According to the fact that the sum of the two sides is greater than the third side, and the difference between the two sides is less than the third side, there is 3-2 < C < 3 + 2, i.e. 1 < C < 5