①a^2 + b^2 + c^2 - ab - ac - bc = 1/2 * (2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc)
= 1/2 [ (a-b)^2 + (b-c)^2 + (a-c)^2 ] = 1/2 * (1+1+4) = 3
② (x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz
則x^2 + y^2 + z^2 = (x+y+z)^2 - 2（xy+yz+xz）= 2^2 - 2*(-5) = 14
③（2+1）（2^2+1）（2^4+1）… （2^64+1）+ 1
= （2-1）（2+1）（2^2+1）（2^4+1）… （2^64+1）+ 1
= （2^2-1）（2^2+1）（2^4+1）… （2^64+1）+ 1
=（2^4-1）（2^4+1）… （2^64+1）+ 1
= ……
= 2^128 - 1 + 1 = 2^128
④ 因?yàn)?x^3 - 5x^2 + 10x - 6 = (x - 1) (x^2 + ax +b) = x^3 - (1 - a)x^2 + (b - a)x - b
所以 1 - a = 5 ,b - a = 10 ,b = 6
所以 a= -4 b = 6
⑤ ( x^2 + nx + 3) ( x^2 - 3x + m) = x^4 + (n-3)x^3 + (m-3n+3)x^2 + (mn-9)x + 3m
依題得 n-3 = 0 m-3n+3 = 0
解得 m = 6 ,n = 3
⑥ 單項(xiàng)式為 am + (a^2)/16 - 1 ( a 不等于 0 )