x^2+2x分之x^2-x+2分之x^2-2x+1÷x+1分之x^2-1
=x²+(x²-x)/2x+(x²-2x+1)/2÷(x²-1)/(x+1)
=x²+(x-1)/2+(x-1)²/2÷(x-1)
=x²+(x-1)/2+(x-1)/2
=x²+x-1
=x²+x+(1/2)²-(1/2)²-1
=(x+1/2)²-5/4
(x+1/2)²-5/4=0
x=±√5/2-1/2=(±√5-1)/2=(√5-1)/2,(-√5-1)/2(不合题意,舍去)
∴x=(√5-1)/2
x^2+2x分之x^2-x+2分之x^2-2x+1÷x+1分之x^2-1=x²+(x²-x)/2x+(x²-2x+1)/2÷(x²-1)/(x+1)=x²+(x-1)/2+(x-1)²/2÷(x-1)=x²+(x-1)/2+(x-1)/2=x²+x-1=x²+x+(1/2)²-(1/2)²-1=(x+1/2)²-5/4当x=0时,原式=-1当x=5时,原式=29∵0<x<5所以-1<原式<29
上面那些答案怎么会有4分之5出来的?要分式化简啊,后面明明是5的平方根--
5的平方根的整数:2∴x<5的平方根的整数,即x<2.当x=2时,原式=5.∵0<x<5所以-1<原式<5.