Question

(1/3)1.若三角形ABC的三边长分别为a、b、c，且满足a^4=b^4+c^4-b^2×c^2,b^4=a^4+c^4-a^2×c^2,c^

Answer 1

三式相加
a^4+b^4+c^4=2a^4+2b^4+2c^4-a^2b^2-a^2c^2-b^2c^2
a^4+b^4+c^4-a^2b^2-b^2c^2-a^2c^2=0
等号两边×2
（a^2-b^2)^2+(b^2-c^2)^2+（c^2-a^2)^2=0
a^2=b^2=c^2
因此为正三角形，即为D

Answer 2

解：
a^4=b^4+c^4-b^2×c^2,
b^4=a^4+c^4-a^2×c^2,
c^4=a^4+b^4-a^2×b^2
三式相加可得
a^4+b^4+c^4-a²b²-a²c²-b²c²=0
(a²-b²)²+(b²-c²)²+(c²-a²)²=2
∴a=b=c
所以是等边三角形
选D

Answer 3

△ABC 是等边三角形。