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

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

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======以下答案可供参考======供参考答案1:[(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供参考答案2: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供参考答案3:[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=1/21/3-1/3(1/2-1)=1供参考答案4: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供参考答案5:[(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

哦，回答的不错

如以上回答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！

点此我要举报以上问答信息