Given that 2A = 3, 2b = 6, 2C = 12, try to judge the relationship between a, B, C

Given that 2A = 3, 2b = 6, 2C = 12, try to judge the relationship between a, B, C

∵ 2A = 3, 2b = 6, 2C = 12, and 6 × 6 = 62 = 3 × 12,
∴（2b）2=2a×2c=2a+c，
∴2b=a+c．

Triangle ABC is 3ABC, box xwyz is - 4x ^ YW ^ Z, and Mn2 is multiplied by nm25

a^3+b^3+c^3=3abc a^3+b^3+c^3-3abc=0 (a+b)^3+c^3-3a^2b-3ab^2-3abc=0 (a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)=0 (a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0 (a+b+c)[a^2+b^2-ac-bc+c^2-ab]=0 (1/2)(a+b+c)[(a-b)^2+(b-c)...

If the three sides of △ ABC satisfy A4 + b2c2-a2c2-b4 = 0, please judge the shape of △ ABC

A4 + b2c2-a2c2-b4 = (a4-b4) + (b2c2-a2c2) = (A2 + B2) (A2-B2) - C2 (A2-B2) = (A2-B2) (A2 + b2-c2) = (a + b) (a-b) (A2 + b2-c2) = 0, ∵ a + b > 0,

It is known that a, B, C are the three sides of the triangle ABC, and a square C square - b square C square = a fourth power - B fourth power, judge the shape of the triangle

a²c²-b²c²=a⁴-b⁴
c²(a²-b²)=(a²+b²)(a²-b²)
A? B? = 0 or C? = a? + B
Namely: triangle ABC is isosceles triangle or right triangle

It is known that a, B, C are the three sides of the triangle ABC, and its shape can be judged by satisfying a square C square - b square C square = the fourth power of a-b Because a square C square - b square C square = a square - b square C square = the fourth power of a - b square, so C square (a square - b square) = (a square + b square) (a square - b square) B step, so C square = a square + B square triangle is a right triangle. Which step is wrong and what is the correct conclusion

Step B is wrong. C? (a? - B?) = (a? + B? (a? - B?) C? (a? B?) - (a? B?) (a? B?) = 0 (a + b) (a-b) (C? - A? A - b) = 0 ν A-B = 0 or C? = a? B &

It is known that A.B.C is the three sides of the triangle ABC, and a square times C square - b square times C square = a fourth power - B fourth power

The original equation is transformed into:
﹙a²c²－b²c²﹚－﹙a^4－b^4﹚=0
∴c²﹙a²－b²﹚－﹙a²＋b²﹚﹙a²－b²﹚=0
∴﹙a²－b²﹚[c²－﹙a²＋b²﹚]=0
 a ﹣ B ﹣ 2 = 0 or C ﹣ 2  0
 a = B or a ﹣ B ﹣ B  C  a = B or a  B  C  a
ν Δ ABC is isosceles △ or right angle △ s

It is known that a, B, C are the three sides of the triangle ABC, and satisfy that a square C-B square C = the fourth power of A-B power

A power C-B power C power = the fourth power of a-the fourth power of B
C (a-b) = (a + b) (a-b)
Party C (a-b) - (a + b) (a-b) = 0
(Party A - Party B) [Party C - (Party A + Party B)] = 0
So a = B or C = a + B
therefore
Triangle ABC may be an isosceles triangle, a right triangle, or an isosceles right triangle

It is known that a, B and C are the three side lengths of the triangle ABC, and the square of a plus the square of 2B plus the square of C minus 2B + (a + C) - 0 can be used to judge the shape of the triangle

(a-b)^2+(b-c)^2=0
∵(a-b)^2>=0,(b-c)^2>=0
∴{(a-b)^2=0,(b-c)^2=0}
∴{a=b,b=c}
The triangle is equilateral

It is known that A.B.C is the three side length of the triangle ABC, and satisfies a square + 2 * b square + C square - 2b (a + C) = 0, try to judge its shape As the title

a²+2b²+c²-2b（a+c）=0
a²+b²-2ab+b²+c²-2bc=0
（a-b）²+（b-c）²）=0
So, A-B = 0, B-C = 0
That is, a = b = C, so the triangle ABC is an equilateral triangle

a. B, C are the lengths of the three sides of the triangle ABC. If the square of a is twice the square of B plus the square of C minus 2B (a + C) = 0, try to judge the shape of the triangle

a^2+2b^2+c^2-2b[a+c]=0
[a^2-2ab+b^2]+[b^2-2bc+c^2]=0
[a-b]^2+[b-c]^2=0
a-b=0===>a=b
b-c=0==>b=c
So: a = b = C
A triangle is an equilateral triangle