It is known that the three sides of the triangle ABC are a, B, C, and satisfy the relation a square + b square + C square + 50 = 6A + 8b + 10C,

It is known that the three sides of the triangle ABC are a, B, C, and satisfy the relation a square + b square + C square + 50 = 6A + 8b + 10C,

a^2-6a+9+b^2-8b+16+c^2-10c+25=0
(a-3)^2+(b-4)^2+(c-5)^2=0
Because square > = 0, the original equation holds only when the formula = 0,
There is a = 3 B = 4 C = 5
Hehe, three strands, four strings and five strings

The three sides of the triangle ABC are a, B, C, and - C square + a square + 2ab-2bc = 0, please explain that triangle ABC is an isosceles triangle

a2+b2+2ab-b2-c2-2bc=0
(a+b)2=(b+c)2
a>0,b>0,c>0
So a + B = C + B
A=c
So it's isosceles

The three side lengths a, B, C of △ ABC satisfy the following relations: - C2 + A2 + 2ab-2bc = 0, try to explain that this triangle is isosceles triangle

(a+c)(a-c)+2b(a-c)=0
(a-c)(a+c+2b)=0
a+c+2b>0
So a-c = 0
A=c
It's an isosceles triangle

Let's say that the square of abc-2ac is an isosceles triangle?

The solution is as follows: - the square of C + the square of a + 2ab-2bc = 0, the shift + 2ab-2bc = the square of C - the square of a - the common factor B is proposed on the left, and 2B (A-C) = (C-A) (c + a) if a-c is not 0, then - 2b = a + C. but we know that the variable length can not be negative, so we can not

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

（a²+b²-c²）²-4a²b²
=（a²+b²-c²+2ab）（a²+b²-c²-2ab）
=[(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

For example: given that ABC is an integer and satisfies the inequality a 2 + B 2 + C 2 + 49 < 4A + 6B + 12C, find the ABC power of the algebraic expression (1 / A + 1 / B + 1 / C)

The problem should be known that ABC is an integer and satisfies the inequality a 2 + B 2 + C 2 + 48 < 4A + 6B + 12C, and find the ABC power of the algebraic formula (1 / A + 1 / B + 1 / C)
a²＋b²＋c²＋48＜4a＋6b＋12c
a²-4a+4+b²-6b+9+c²-12c+36＜1
(a-2)²+(b-3)²+(c-6)²＜1
∵ ABC is an integer
∴a=2 b=3 c=6
The ABC power of (1 / A + 1 / B + 1 / C)
=The 36th power of (1 / 2 + 1 / 3 + 1 / 6)
=The 36th power of 1
=1

Given that the three sides of the unequal edge △ ABC are integers a, B, C respectively, and satisfy the square of a + the square of B - 4a-6b + 13, find the length of C

The sum of the two sides of the triangle is greater than the third side and the difference between the two sides of the triangle is less than the third side

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

Given that positive integers a, B, C satisfy the inequality: A ^ 2 + B ^ 2 + C ^ 2 + 49 ≤ 4A + 6B + 12C, find the value of (1 / A + 1 / B + 1 / C) ^ ABC

a^2+b^2+c^2+49≤4a+6b+12c
（a-2）²+（b-3)²+（c-6）²≤0
So a = 2, B = 3, C = 6
And then it's good to calculate 1 / A + 1 / B + 1 / C = 1 / 2 + 1 / 3 + 1 / 6 = 1
The result is 1

If a, B, C are the three sides of △ ABC, and A2 + B2 + C2 = AB + AC + BC, then △ ABC is () A. Isosceles triangle B. Right triangle C. Equilateral triangle D. Isosceles right triangle

The original formula can be changed into 2A2 + 2B2 + 2c2 = 2Ab + 2Ac + 2BC, that is, A2 + B2 + C2 + A2 + B2 + c2-2ab-2ac-2bc = 0;
According to the complete square formula, we get: (a-b) 2 + (C-A) 2 + (B-C) 2 = 0;
From the properties of nonnegative numbers, we can know that A-B = 0, C-A = 0, B-C = 0; that is, a = b = C. Therefore, △ ABC is an equilateral triangle
Therefore, C