In the triangle ABC, the length of the three sides is ABC, the square of a = n minus one, B = 2n, the square of C = n plus one of this triangle

What is the requirement?
B = 2n, the three sides are not the same power of n
If, B = 2n squared
Then, the square of a + the square of C = the square of B
B is the hypotenuse, a and C are right angles
Triangle ABC is a right triangle

It is known that in the triangle ABC, the length of the three sides is ABC, and a is equal to N, and B is equal to 4 / n square minus 1. Please prove that the triangle is a right triangle, please It is known that in the triangle ABC, the three sides are ABC respectively, and a is equal to N, B is equal to 4 / n square minus 1, C is equal to 4 / n square plus 1, [n is an even number greater than 2], it is proved that the triangle is a right triangle

Because n is an even number greater than 2, n ^ 2 / 4 + 1 > N and n ^ 2 / 4 + 1 > n ^ 2 / 4-1, so n ^ 2 / 4 + 1 is the longest side. To prove a right triangle, prove n ^ 2 + (n ^ 2 / 4-1) ^ 2 = (n ^ 2 / 4 + 1) ^ 2.n ^ 2 + n ^ 4 / 16-n ^ 2 / 2 + 1 = n ^ 4 / 16 + n ^ 2 / 2 + 1 N ^ 2-N ^ 2 / 2 = n ^ 2 / 2, so this triangle is a right triangle

The three sides of the triangle ABC are a, B, C. If a square plus b square plus C square is equal to ab plus AC and CB, try to judge the shape of triangle ABC Very urgent. Hand in your homework tomorrow Use factorization to do it. Please write in detail and understand more

∵ a  a ᙽ B ᙽ C ᙽ a ᙽ a ᙽ a ᙽ B  C ᙽ a ᙽ a ᙽ B ? B ? C ? a ? B ? B ᙾ C ? a  B ᙽ C ? a  B ? B ? B

When AB + BC is equal to the square of B + AC, the shape of triangle ABC is judged

ab+bc=b^2+ac
ab-ac=b^2-bc
a(b-c)=b(b-c)
(a-b)(b-c)=0
A = B or B = C
So the triangle ABC must be an isosceles triangle

The three side lengths of △ ABC a, B, C satisfy a + B = 8, ab = 4, C2 = 56. Try to judge the shape of △ ABC and explain the reasons

∵a+b=8，
∴（a+b）2=64，
a2+b2+2ab=64，
∴ab=4，
∴a2+b2=64-2ab=64-8=56=c2，
The shape of △ ABC is a right triangle

The shape of the triangle ABC can be judged if a square minus AC is equal to b square minus BC speed

a²-ac-b²+bc=0
(a-b)(a+b)-c(a-b)=0
(a-b-c)(a-b)=0
∵a-b-c

It is known that in the triangle ABC, a, B, C are three sides, and the square of a + the square of B + the square of C - AB AC BC = 0, indicating that the triangle ABC is an equilateral triangle I just on the first day of junior high school, the answer should be suitable for the second volume of our junior high school mathematics book (Suke Edition). I saw someone answer this in Baidu: square of a + square of B + square of C - AB AC BC= =1 / 2 (the square of a-2ab + the square of B + the square of b-2bc + the square of C + the square of a-2ac + the square of C) =1/2[(a-b)^2+(b-c)^2+(c-a)^2]=0 a-b=0,b-c=0c-a=0 a=b=c The triangle ABC is an equilateral triangle 1 / 2 [(a-b) ^ 2 + (B-C) ^ 2 + (C-A) ^ 2] = 0 What does this inverted V mean? Help me do it directly!

It seems that you should have been able to do it
（a-b)^2=(a-b)²
Because power is not easy to express, we sometimes use ^ to express several powers

If ABC is the three sides of a triangle, and the square of a + the square of B + the square of C = AB + BC + Ca, please explain that the triangle ABC is an equilateral triangle

∵a²+b²+c²=ab+ac+bc
∴a²+b²+a²+c²+b²+c²=2ab+2ac+bc
∴(a-b)²+(a-c)²+(b-c)²=0
∴a=b=c
⊿ ABC is an equilateral triangle

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

0

Because a square + b square + C square = AB + BC + ca
Both sides of the equation * 2
Results: 2A square + 2B square + 2C square - 2ab-2bc-2ca = 0
Deformation: (a-b) square + (A-C) square + (B-C) square = 0
Because all three sides are positive real numbers, we deduce a = b = C
So it's an equilateral triangle

In the triangle ABC, the length of the three sides is ABC, the square of a = n minus one, B = 2n, the square of C = n plus one of this triangle