標題:
二項式問題
發問:
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1)已知x+=3，y+z=2，求x^2+y^2+z^2+xy+yz-zx的值 答案:7 2)已知x-y=2，x^2+y^2+z^2-xy-yz-zx=19，求y-z的最大值 答案:3
最佳解答:
(1) If x+y = 3, y+z = 2 Then (y+z)-(x+y) = 2 - 3 z - x = -1 x^2 + y^2 + z^2 + xy + yz - zx = 1/2 (2x^2 + 2y^2 + 2z^2 + 2xy + 2yz - 2zx) = 1/2 (x^2 + 2xy + y^2 + z^2 - 2zx + x^2 + y^2 + 2yz + z^2) = 1/2 [ (x+y)^2 + (z-x)^2 + (y+z)^2 ] = 1/2 [ 3^2 + (-1)^2 + 2^2 ] = 1/2 (9+1+4) = 7 (2) x^2 + y^2 + z^2 - xy - yz - zx = 19 2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx = 38 (x^2 - 2xy + y^2) + (z^2 - 2zx + x^2) + (y^2 - 2yz + z^2) = 38 (x-y)^2 + (z-x)^2 + (y-z)^2 = 38 2^2 + (z-x)^2 + (y-z)^2 = 38 (z-x)^2 + (y-z)^2 = 34 ................(*) Let y-z = M Then z-x = -(y-z)-(x-y) = -M-2 Put into (*) (-M-2)^2 + M^2 = 34 M^2 + 4M + 4 + M^2 = 34 M^2 + 2M + 2 = 17 用完成平方法 (By completing the square) (M + 1)^2 + 1 = 17 (M + 1)^2 = 16 (M + 1) = 4 or -4 M = 3 or -5 因為 M = y - z 所以，y - z 的最大值為 3
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