(1)2(2x+1)-(x+5)-2x-3=x+1(2)单项式3a∧x-1 b∧6与-1/2a∧2(x-3) b∧2+y之差是单项式,试求x,y的值.题中空格为不是次方.

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(1)2(2x+1)-(x+5)-2x-3=x+1(2)单项式3a∧x-1 b∧6与-1/2a∧2(x-3) b∧2+y之差是单项式,试求x,y的值.题中空格为不是次方.
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(1)2(2x+1)-(x+5)-2x-3=x+1(2)单项式3a∧x-1 b∧6与-1/2a∧2(x-3) b∧2+y之差是单项式,试求x,y的值.题中空格为不是次方.
(1)2(2x+1)-(x+5)-2x-3=x+1
(2)单项式3a∧x-1 b∧6与-1/2a∧2(x-3) b∧2+y之差是单项式,试求x,y的值.
题中空格为不是次方.

(1)2(2x+1)-(x+5)-2x-3=x+1(2)单项式3a∧x-1 b∧6与-1/2a∧2(x-3) b∧2+y之差是单项式,试求x,y的值.题中空格为不是次方.
x>-1,求y=(x+5)(x+2)/(x+1)的最小值?
y=(x+5)(x+2)/(x+1)=(x^2+7x+10)/(x+1)
y(x+1)=x^2+7x+10
x^2+(7-y)x+10-y=0………………(1)
∵x∈R,∴其判别式△=(7-y)^2-4(10-y)=y^2-14y+49-40+4y
=y^2-10y+9=(y-1)(y-9)≥0
得y≤1或y≥9.
将y=9代入(1)式,得x^2-2x+1=(x-1)^2=0,于是得x=1>-1.
将y=1代入(1)式,得x^2+6x+9=(x+3)^2=0,于是得x=-3-1,∴当x=1时获得y的最小值ymin=9

(1)x=0
(2)x=8,y=4

不可能,得数是 X-6=X+1

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