Question

已知三角形的边为a,b,c,判断a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2的值.

Answer 1

它的值小于0,理由如下：

a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2

=(a^4-2a^2b^2+b^4)-2a^2c^2+2b^2c^2+c^4-4b^2c^2

=(a^2-b^2)^2-2c^2*(a^2-b^2)+c^4-4b^2c^2

=(a^2-b^2-c^2)^2-4b^2c^2

=(a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)

=[a^2-(b^2-2bc+c^2)][a^2-(b^2＋2bc＋c^2)]

=[a^2-(b－c)^2][a^2-(b＋c)^2]

=(a+b-c)(a-b+c)(a+b+c)(a-b-c)

可知：a+b-c>0,a-b+c>0,a+b+c>0,a-b-c