a^4+b^4+c^4=a^2b^2+b^2c^2+a^2c^2两边同时乘2,得：2a^4+2b^4+2c^4=2a^2b^2+2b^2c^2+2a^2c^2(a^4-2a^2b^2+b^4)+(a^4-2a^2c^2+c^4)+(b^4-2b^2c^2+c^4)=0(a^2-b^2)^2+(a^2-c^2)^2+(b^2-c^2)^2=0平方数都是非负数三个非...(a-2b+c+3d)(a+2b-c+3d(-2/3^m-3/5^2)^2(b+c)(-b-c)以后请单独发，不用加悬赏你这种追问，和原题无关(a-2b+c+3d)(a+2b-c+3d)=[(a+3d)-(2b-c)]*[(a+3d)+(2b-c)]=(a+3d)²-(2b-c)²=a²+6d+9d²-4b²+4bc-c²(-2/3^m-3/5^2)^2太乱，看不懂(-2/3^m-3/5^2)^2=(2/3^m+3/5^2)^2=4/9^m+12/(25*3^m)+9/625按你列的式子，应该是这样，(b+c)(-b-c)=b²-c²第二道打错了，我我重打(-2/3m^2-3/5n^2)^2(-2/3m^2-3/5n^2)^2=(2/3m^2+2/5n^2)^2=3/9m^4+2*8/15m^2*n^2+4/25n^4