Question

小括号x-3分之2x-2在负1括号完除以x开方-9分之x平方+2x+1,其中x等于负2分之1

Answer 1

解：原式=[(2x-2)/(x-3)-1]÷[(x²+2x+1)/(x²-9)]
=[(x+1)/(x-3)]÷[(x+1)²/(x+3)(x-3)]
=[(x+1)/(x-3)]×[(x+3)(x-3)/(x+1)²]
=(x+3)/(x+1)
当x=-1/2时
原式=(x+3)/(x+1)
=(-1/2+3)(-1/2+1)
=5/4

Answer 2

[(2x-2)/(x-3)-1]÷[x-√(x²+2x+1)/9]
=[(x+1)/(x-3)]÷[x-|x+1|/3]
∵x=-1/2
∴原式=[(1/2)/(-7/2)]÷[-1/2-(1/2)/3]
=(-1/7)÷（-1/2-1/6）
=(-1/7)*(-6/4)
=3/14

Answer 3

你能用式子表达出来吗？这样写让人不太明白你写的是什么。