1已知│x-y+1│与x2+8x+16互为相反数,求x2+2xy+y2的值

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1已知│x-y+1│与x2+8x+16互为相反数,求x2+2xy+y2的值
2已知2a-3x+1=0,3b-2x-16=0,且a小于、等于4,4小于b,求x的取值范围.
3当n为正整数时,代数式n(n+7)-(n-3)(n-2)的值是否能被6整除?理由?

(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

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