初中数学先化简再求值
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解：
原式=2/(x-1)-[1/(x²-1)]×[(x²+2x+1)/(x-1)]
=2/(x-1)-{1/[(x+1)(x-1)]}×[(x+1)²/(x-1)]
=2/(x-1)-[(x+1)/(x-1)²]
=2(x-1)/(x-1)²-[(x+1)/(x-1)²]
=[2(x-1)-(x+1)]/(x-1)²
=(2x-2-x-1)/(x-1)²
=(x-3)/(x-1)²
=(√3+1-3)/(√3+1-1)²
=(√3-2)/(√3)²
=(√3-2)/3
原式=2/(x-1)-[1/(x²-1)]×[(x²+2x+1)/(x-1)]
=2/(x-1)-{1/[(x+1)(x-1)]}×[(x+1)²/(x-1)]
=2/(x-1)-[(x+1)/(x-1)²]
=2(x-1)/(x-1)²-[(x+1)/(x-1)²]
=[2(x-1)-(x+1)]/(x-1)²
=(2x-2-x-1)/(x-1)²
=(x-3)/(x-1)²
=(√3+1-3)/(√3+1-1)²
=(√3-2)/(√3)²
=(√3-2)/3
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2/(x-1)-1/(x^2-1) (x^2+2x+1)/(x-1)
=2/(x-1)-1/((x+1) (_(x-1))) 〖(x+1)〗^2/((x-1))
=2/(x-1)-(x+1)/〖(x-1)〗^2
= (x-3)/〖(x-1)〗^2
代入x=√3+1得原式=(√3-2)/3
=2/(x-1)-1/((x+1) (_(x-1))) 〖(x+1)〗^2/((x-1))
=2/(x-1)-(x+1)/〖(x-1)〗^2
= (x-3)/〖(x-1)〗^2
代入x=√3+1得原式=(√3-2)/3
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原式=2/(x-1)-[1/(x²-1)]×[(x²+2x+1)/(x-1)]
=2/(x-1)-{1/[(x+1)(x-1)]}×[(x+1)²/(x-1)]
=2/(x-1)-[(x+1)/(x-1)²]
=2(x-1)/(x-1)²-[(x+1)/(x-1)²]
=[2(x-1)-(x+1)]/(x-1)²
=(2x-2-x-1)/(x-1)²
=(x-3)/(x-1)²
因为x√3+1，
所以上式=(√3+1-3)/(√3+1-1)²
=(√3-2)/(√3)²
=(√3-2)/3
=2/(x-1)-{1/[(x+1)(x-1)]}×[(x+1)²/(x-1)]
=2/(x-1)-[(x+1)/(x-1)²]
=2(x-1)/(x-1)²-[(x+1)/(x-1)²]
=[2(x-1)-(x+1)]/(x-1)²
=(2x-2-x-1)/(x-1)²
=(x-3)/(x-1)²
因为x√3+1，
所以上式=(√3+1-3)/(√3+1-1)²
=(√3-2)/(√3)²
=(√3-2)/3
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