热心网友
∵5a^2+b^2+c^2=(2√３)ac+(２√２)ab∴5a^2+b^2+c^2-(2√３)ac-(２√２)ab=[3a^2-(2√３)ac+c^2]+[2a^2-(２√２)ab+b^2]=[(√３)a-c]^2 + [(√2)a-b]^2 =0即c=(√３)a；b=(√2)a很明显：c^2=b^2+a^2∴三角形是以C为直角Rt三角形
热心网友
5a^2加b^2加c^2=2倍根号３乘以ac加２倍根号２乘ab3a^2-2倍根号３乘以ac+c^2+b^2-２倍根号２乘ab+2a^2=0(根号３乘以a-c)^2+(根号２乘a-b)^2=0根号３乘以a=c,根号２乘a=ba^2+b^2=c^2所以是直角三角形