先化简再求值[[(x2+2x+1)/(x+2)]÷[(x2-1)/(x-1)]-1/(x+2)其中x=√2-2
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先化简再求值[[(x2+2x+1)/(x+2)]÷[(x2-1)/(x-1)]-1/(x+2)其中x=√2-2先化简再求值[[(x2+2x+1)/(x+2)]÷[(x2-1)/(x-1)]-1/(x+
先化简再求值[[(x2+2x+1)/(x+2)]÷[(x2-1)/(x-1)]-1/(x+2)其中x=√2-2
先化简再求值[[(x2+2x+1)/(x+2)]÷[(x2-1)/(x-1)]-1/(x+2)其中x=√2-2
先化简再求值[[(x2+2x+1)/(x+2)]÷[(x2-1)/(x-1)]-1/(x+2)其中x=√2-2
解
原式
=[(x+1)²/(x+2)]×[(x-1)/(x-1)(x+1)]-1/(x+2)
=[(x+1)²/(x+2)]×1/(x+1)-1/(x+2)
=(x+1)/(x+2)-1/(x+2)
=x/(x+2)
=(√2-2)/(√2-2+2)
=(√2-2)/(√2)
=1-√2
原式=(x+1)²/(x+2)÷(x+1)(x-1)/(x-1)-x/(x+2)
=(x+1)²/(x+2)÷(x+1)-x/(x+2)
=(x+1)/(x+2)-x/(x+2)
=(x+1-x)/(x+2)
=1/(x+2)
=1/(√3-2+2)
=√3/3
求采纳为满意回答。
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