If the three sides of the triangle ABC are a, B, C, and satisfy the condition that the cubic power of a - the square of a B + the square of ab - the square of AC + the cubic power of BC - B = 0, find the shape of the triangle ABC

If the three sides of the triangle ABC are a, B, C, and satisfy the condition that the cubic power of a - the square of a B + the square of ab - the square of AC + the cubic power of BC - B = 0, find the shape of the triangle ABC

a^3-a²b+ab²-ac²+bc²-b^3=0
(a^3-b^3)-(a²b-ab²)-(ac²-bc²)=0
(a-b)(a²+ab+b²)-ab(a-b)-c²(a-b)=0
(a-b)(a²+ab+b²-ab-c²)=0
(a-b)(a²+b²-c²)=0
So (a-b) = 0 or (a? + B? - C?) = 0
So, a = B or a? + B? = C
So it's an isosceles triangle or a right triangle

If A.B.C is the three sides of the triangle ABC and satisfies the quadratic power of a + the quadratic power of B + the quadratic power of C - AB BC CA = 0

The second power of a + the second power of B + the second power of C - AB BC CA = 02A 2 + 2B 2 + 2C 2 - 2 ab - 2 BC - 2 CA = 0 (a 2 - 2 ab + B 2) + (B 2 - 2 BC + C 2) + (C 2 - 2 Ca + a 2) = 0 (a - b) 2 + (B - C) 2 + (C - a) 2) = 0, so a - B = 0, B - C = 0

If a, B, C are three sides of △ ABC and satisfy the condition a ^ 2 + B ^ 2 + C ^ 2 = AB + BC + Ca, what kind of triangle is this triangle

Multiply both sides by 2
2a2＋2b2＋2c2＝2ab＋2bc＋2ca
2a2＋2b2＋2c2－2ab－2bc－2ca＝0
（a－b）2＋（b－c）2＋（c－a）2＝0
∵（a－b）2≥0 ,（b－c）2≥0 ,（c－a）2≥0 ,∴ a－b＝0 ,b－c＝0 ,c－a＝0
A = b = C, equilateral triangle

Let a, B, C of the triangle ABC satisfy the quadratic power of a + the second power of B + the second power of C = AB + BC + CD

Equilateral triangle
a²+b²+c²=1/2（（a²+b²）+（a²+c²）+（b²+c²））
a²+b²≥2ab a²+c²≥2ac,b²+c²≥2bc
So a 2 + B 2 + C 2 ≥ 1 / 2 (2Ab + 2BC + 2Ac) = AB + BC + AC
It holds when a = b = C, so it is an equilateral triangle

If the side length of the triangle ABC is a, B, C. if the cubic power of a - the square of a B + the square of ab - the square of AC + the cubic power of BC - B = 0, the shape of triangle ABC is Please! It's due tomorrow. Thank you!

a^3-a²b+ab²-ac²+bc²-b^3=0
(a^3-b^3)-(a²b-ab²)-(ac²-bc²)=0
(a-b)(a²+ab+b²)-ab(a-b)-c²(a-b)=0
(a-b)(a²+ab+b²-ab-c²)=0
(a-b)(a²+b²-c²)=0
So (a-b) = 0 or (a? + B? - C?) = 0
So, a = B or a? + B? = C
So it's an isosceles triangle or a right triangle

a. B. C is the three sides of the triangle ABC, and satisfies the quadratic power of a + the quadratic power of B + the quadratic power of C - AB BC Ca, a new form of triangle ABC

a. B. C is the three sides of the triangle ABC, and satisfies the quadratic power of a + the quadratic power of B + the quadratic power of C - AB BC CA = 0
So there are
2 (the power of a + the power of B + the power of C - AB BC CA)
=The quadratic power of (a-b) + (B-C) + (C-A)
=0
A-B = 0, B-C = 0, C-A = 0
The shape of triangle ABC is an equilateral triangle

1 / 2013 + 2 / 2013 + 3 / 2013 + 2011 / 2013 + 2012 =? I know (2013-1) = 2012 / 2 = 1006

1% in 2013 + 2% in 2013 + 3% in 2013 + 2011 in 2013 + 2012 in 2013
=（1+2+.+2012）/2013
=(1+2012)*2012/2/2013
=2012/2
=1006

2013*2013-2013*2012-2011*2012+2012*2012 Common factor method

2013*2013-2013*2012-2011*2012+2012*2012
=2013(2013-2012)+2012(2012-2011)
=2013*1+2012*1
=2013+2012
=4025
If you agree with my answer, please click "adopt as satisfactory answer" and wish you progress!

It is known that a + x 2 = 2011, B + x 2 = 2012, C + x 2 = 2013, and ABC = 24. Calculate the value of a / BC + C / AB + B / AC = 1 / A-1 / B-1 / C Find the value of a / BC + C / AB + B / ac-1 / A-1 / B-1 / C

It's 1 / 8. Just take ABC = 24 inside

(1) Known: a + x 2 = 2011, B + x 2 = 2012, C + x 2 = 2013, and abv = 24, find a / BC + C / AB + B / ac-1 / A-1 / B-1 / C

Abv = 24, what is v? If ABC = 24, the final result should be 1 / 8