先将分式(x分之x-1减x+1分之x-2)除以x的平方+2x+1分之2x的平方-x化简,再

2025-02-24 03:45:56
推荐回答(2个)
回答1:

(x分之x-1减x+1分之x-2)除以x的平方+2x+1分之2x的平方-x
=[(x-1)/x -(x-2)/(x+1) ]×(x+1)²/x(2x-1)
=[(x-1)(x+1)-x(x-2)]/x(x+1)×(x+1)²/x(2x-1)
=(x²-1-x²+2x)/x²(2x-1)
=(2x-1)/x²(2x-1)
=1/x²
当x=1时
原式=1/1=1

回答2:

原式=[(x-1)/x -(x-2)/(x 1) ]/(x 1)²/x(2x-1)
={[(x-1)(x 1)-x(x-2)]/x(x 1)}*(x 1)²/x(2x-1)
=x-1/x²
当x=1时
原式=0