（x-3）分之x减去（x²-3x）分之x+6加上x分之1

（x-3）分之x减去（x²-3x）分之x+6加上x分之1
（x-3）分之x减去（x²-3x）分之x+6加上x分之1

（x-3）分之x减去（x²-3x）分之x+6加上x分之1

写得有点……希望能帮到你,↖(^ω^)↗

x/(x-3)-(x+6)/(x^2-3x)+1/x =[x^2-(x+6)+(x-3)]/x(x-3) =(x^2-9)/x(x-3) =(x+3)/x。

原式={x²-（x+6）+（x-3）} / x（x-3）
=（x²-9）/x（x-3）
=（x+3）（x-3）/x（x-3）
=（x+3）/x
详细过程给你咯
望采纳哦！~~~~

（x-3）分之x减去（x²-3x）分之x+6加上x分之1
=x/(x-3)-(x+6)/[x(x-3)]+1/x
=(xx-x-6+x-3)/[x(x-3)]
=(xx-9)/[x(x-3)]
=(x-3)(x+3)/[x(x-3)]
=(x+3)/x
=1+3/x

x-3）分之x减去（x²-3x）分之x+6加上x分之1
=（x²-3x）分之x²减去（x²-3x）分之x+6加上（x²-3x）分之（x-3）
=（x²-3x）分之（x²-9）
=x分之（x+3）

x/(x-3)-(x+6)/(x^2-3x)+1/x = (x*x)/(x*(x-3)) - (x+6)/(x*(x-3)) + (x-3)/(x*(x-3)) = (x*x-x-6+x-3)/(x*(x-3)) = (x*x-9)/(x*(x-3)) = (x+3)/x

x/(x-3)-(x+6)/(x^2-3x)+1/x
=x^2/[x*(x-3)]-(x+6)/[x*(x-3)]+(x-3)/[x*(x-3)]
=[x^2-(x+6)+(x-3)]/[x*(x-3)]
=(x^2-x-6+x-3)/[x*(x-3)]
=(x^2-9)/[x*(x-3)]
=(x+3)*(x-3)/[x*(x-3)]
=(x+3)/x
=1+3/x