It is known that the three sides of the triangle ABC satisfy a = B + 1, ab = 12, C = 5. Is triangle ABC a right triangle? Proof judgment

It is known that the three sides of the triangle ABC satisfy a = B + 1, ab = 12, C = 5. Is triangle ABC a right triangle? Proof judgment

yes
Because B = 1
So AB = 12
b(b+1)=12
B=3
A=4
According to Pythagorean theorem:
yes

Given that the three sides of the triangle ABC are a, B, C, and the three sides satisfy the equation a ^ 2 + B ^ 2 + C ^ 2 = AB + BC + AC, try to judge the shape of the triangle ABC

A ^ 2 + B ^ 2 + C ^ 2 = AB + BC + ACA? + B? + C? - AB BC CA = 02 (a? + B? + C? - AB BC CA) = 0A ^ 2-2ab + B ^ 2 + B ^ 2 + B ^ 2 + C ^ 2-2ac + A ^ 2 = 0 (a-b) ^ 2 + (B-C) ^ 2 + (C-A) ^ 2 = 0, so a = b = C is an equilateral triangle

Given the three sides of a triangle a, B, C fit the equation: A3 + B3 + C3 = 3ABC, please determine the shape of the triangle

A3 + B3 + C3 + C3 = 3ABC, and A3 + B3 + C3 + c3-3abc + a3 + B3 + C3 + C3 + c3-3abc = (a + b) (a2-ab + B2) + c3-3abc = (a + b) [(a + b) [(a + b) 2-3ab] + c3-3abc = (a + b) 3-3ab (a + b) 3-3ab (a + b) + c3-3abc = (a + b) 3-3ab (a + b) a + B + C 3-3abc = (a + b) [(a + b) 2 - (a + b) C + C + C2] - 3AB (a + B + C + A + B + C + A + B (a + B + C) = (a + B + B + C) = (a + B + B + c+ C) (A2 + B2 + C2 AB AC BC) = 0 ∵

If A.B.C is the three sides of the triangle ABC, and the square of a + the square of B + the square of C = AB + AC + BC, try to judge the shape of the triangle ABC

a^2+b^2+c^2=ab+ac+bc
2(a^2+b^2+c^2)-2(ac+ab+cb)=0
(a-b)^2+(a-c)^2+(b-c)^2=0
|a-b|=0,a=b
|a-c|=0,a=c
|b-c|=0,b=c
a=b=c
So, it's an equilateral triangle

It is known that a, B, C are the three sides of the triangle ABC, and the square of a + the square of B + the square of C - AB BC AC = 0, please judge the shape of triangle ABC Why?

The square of a + the square of B + the square of C - AB BC - AC = 0, so 2 (the square of a + the square of B + the square of C - AB BC AC) = 0 (both sides are multiplied by 2) the square of 2A + the square of 2B + the square of 2C - 2ab-2bc-2ac = 0 can match the square of a - 2Ab + the square of B + the square of B - 2

It is known that ABC is the three side length of a triangle ABC, and satisfies the square of a + the square of B + the square of C - AB BC AC = 0

Equilateral triangle
a²+b²+c²-ab-bc-ac=0
a(a-b)+b(b-c)+c(c-a)=0
Because a, B, C are all greater than 0
So the condition that the above equation equals zero is
a=b;b=c;c=a
That is, a = b = C
Therefore, the triangle is equilateral

Given that the three sides of a triangle a, B, C satisfy A2 + B2 + C2 = AB + BC + Ca, then the shape of the triangle () A. Right triangle B. Isosceles triangle C. Equilateral triangle D. There is a right triangle with an angle of 30 degrees

∵a2+b2+c2=ab+bc+ca
Multiply both sides by 2 to get: 2A2 + 2B2 + 2c2-2ab-2bc-2ac = 0
That is (a2-2ab + B2) + (b2-2bc + C2) + (c2-2ac + A2) = 0
∴（a-b）2+（b-c）2+（c-a）2=0
∵ even power is always greater than or equal to 0,
∴a-b=0，b-c=0，c-a=0
∴a=b，b=c，c=a．
So this is an equilateral triangle
Therefore, C

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

It is known that the trilateral a, B and C of △ ABC satisfy the equation A2 + B+| c−1-2|=6a+2 B − 3-7, try to judge the shape of △ ABC

∵a2+b+|
c−1-2|=6a+2
b−3-7，
∴a2+b+|
c−1-2|-6a-2
b−3+7=0，
∴a2-6a+9+[（b-3）-2
b−3+1]+|
c−1-2|=0，
(A-3) 2+（
b−3-1）2+|
c−1-2|=0，
∴a=3，b=4，c=5，
∵32+42=52，
The triangle is a right triangle

If the triangle ABC satisfies the equation a ^ 2 + B + L √ C-1 - 2L = 6A + 2 √ B-3 - 7, a good additional score is obtained to judge the shape of triangle ABC The triangle ABC satisfies the equation a ^ 2 + B + L √ C-1 - 2L = 6A + 2 √ B-3 - 7 to judge the shape of triangle ABC OK, additional points Answer "blulluee": A ^ 2 + B + L (radical C-1) - 2L = 6A + 2 times (radical B-3) - 7 I added that bracket myself, so that you can see it clearly "L l l" is an absolute value

a^2+b+l √(c-1)-2l=6a+2√(b-3) -7a^2-6a+b+l √(c-1)-2l=2√(b-3) -7a^2-6a+b-2√(b-3) +l √(c-1)-2l+7= 0a^2-6a+b-3-2√(b-3) +l √(c-1)-2l+7+3= 0a^2-6a+b-3-2√(b-3) +l √(c-1)-2l+10= 0a^2-6a+9+b-3-2√(b-...