这道分式题怎么做?急!!!!
8个回答
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x/(x-1)-3/(x-1)(x+2)-1
=x(x+2)/(x-1)(x+2)-3/(x-1)(x+2)-(x-1)(x+2)/(x-1)(x+2)
=[x(x+2)-3-(x-1)(x+2)]/(x-1)(x+2)
=[x^2+2x-3-(x-1)(x+2)]/(x-1)(x+2)
=[(x+3)(x-1)-(x-1)(x+2)]/(x-1)(x+2)
=(x-1)[(x+3)-(x+2)]/(x-1)(x+2)
=(x-1)(x+3-x-2)/(x-1)(x+2)
=(x-1)/(x-1)(x+2)
=1/(x+2)
=1/(-2/3+2)
=1/(4/3)
=3/4
=x(x+2)/(x-1)(x+2)-3/(x-1)(x+2)-(x-1)(x+2)/(x-1)(x+2)
=[x(x+2)-3-(x-1)(x+2)]/(x-1)(x+2)
=[x^2+2x-3-(x-1)(x+2)]/(x-1)(x+2)
=[(x+3)(x-1)-(x-1)(x+2)]/(x-1)(x+2)
=(x-1)[(x+3)-(x+2)]/(x-1)(x+2)
=(x-1)(x+3-x-2)/(x-1)(x+2)
=(x-1)/(x-1)(x+2)
=1/(x+2)
=1/(-2/3+2)
=1/(4/3)
=3/4
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3/40
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2011-03-23
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第一个分式分子分母乘(x-1)第三个分子分母乘(x-1)(x+2) 再分式相加 上面的分子是x(x-1)(x+2)-3-(x-1)(x+2) 拆开再因式分解得(x-1)(x+2)(x+3)和分母相约就可以了
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x/(x-1)-3/(x-1)(x+2)-1
=x*(x+2)/[(x-1)(x+2)]-3/(x-1)(x+2)-1
=(x^2+2x-3)/(x-1)(x+2)-1
=(x+3)/(x+2)-1
=3/4
=x*(x+2)/[(x-1)(x+2)]-3/(x-1)(x+2)-1
=(x^2+2x-3)/(x-1)(x+2)-1
=(x+3)/(x+2)-1
=3/4
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等于1吧....
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通分
(x^2+2x-3)/(x-1)(x+2) -1
分子提出x-1消去 剩下的不写了
(x^2+2x-3)/(x-1)(x+2) -1
分子提出x-1消去 剩下的不写了
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