初二数学：先化简,再求值： [(x²-2x)/(x²-1)]/[x-1-(2x-1)/(x+1)]

初二数学：先化简,再求值： [(x²-2x)/(x²-1)]/[x-1-(2x-1)/(x+1)],其中,x=1/2.

binglizhi 1年前已收到6个回答举报

共回答了10个问题采纳率：80% 举报

分母[x-1-(2x-1)/(x+1)]=(x²-1-2x+1))/(x+1)=(x²-2x)/(x+1)[(x²-2x)/(x²-1)]/[x-1-(2x-1)/(x+1)]=[(x²-2x)/(x²-1)]/[(x²-2x)/(x+1)]=1/(X-1)=-2

1年前

wuchi1 幼苗

共回答了31个问题举报

[(x²-2x)/(x²-1)]/[x-1-(2x-1)/(x+1)]
=[(x²-2x)/(x-1)]/(x²-2x) (分子分母同乘x+1)
=1/(x-1)
=-2

1年前

共回答了4个问题举报

分子=x(x-2)/(x+1)(x-1)
分母通分,=[x^2-1-(2x-1)]/(x+1)=(x^2-2x)/(x+1)=x(x-2)/(x+1)
最后=1/(x-1) 代入数值=-2

1年前

共回答了568个问题举报

[(x²-2x)/(x²-1)]÷[x-1-(2x-1)/(x+1)],
=[x(x-2)]/[(x+1)(x-1)]÷[x^2-1-2x+1[/(x-1)
=[x(x-2)]/[(x+1)(x-1)]×(x-1)/[x(x-2)]
=1/(x+1)
=1/(1/2+1)=2/3

1年前

共回答了147个问题举报

原式=[x(x-2)/(x^2-1)]/[(x^2-1-2x+1)/(x+1)]=[x(x-2)/(x^2-1)]/[(x^2-2x)/(x-1)]=(x-1)/(x^2-1)=1/(x+1)=2/3

1年前

共回答了7个问题举报

[x(x-2)/(x+1)(x-1)]/[3x/(x+1)]=[x(x-2)/(x+1)(x-1)] [(x+1)/3x]=(x-2)/3(x-1)=1/3-1/3(x-1)
x=1/2
1/3-1/3(1/2-1)=1

1年前

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初二数学 ：先化简,再求值： [(x²-2x)/(x²-1)]/[x-1-(2x-1)/(x+1)]