Question

1已知│x-y+1│与x2+8x+16互为相反数,求x2+2xy+y2的值

Answer 1

(1) |x-y+1| ≥0,x^2+8x+16=(x+4)^2≥0

两者互为相反数,所以|x-y+1| ＝ x^2+8x+16=0

x-y+1=0,x+4=0,所以x=-4,y=-3

x2+2xy+y2 = (x+y)^2 = (-7)^2 = 49

(2) 2a-3x+1=0,3b-2x-16=0

所以 x= (2a+1)/3 = (3b-16)/2

a≤4,所以 x ≤ (2*4+1)/3=3

b>4,所以 x > (3*4-16)/2 = -2

综合一下得到,-2