1.计算:a/(x^2+ax)-a/(x^2-a^2) 2.解方程:x/(x-5)-1=5/(x^2-10x+25)
1个回答

1.a/(x^2+ax)-a/(x^2-a^2) =a/x(x+a) - a/(x+a)(x-a)=(ax-a²-ax)/x(x+a)(x-a)= -a²/x(x+a)(x-a)

2.x/(x-5)-1=5/(x^2-10x+25)

(x-x+5)/(x-5)=5/(x-5)²

5/(x-5)=5/(x-5)²

1=1/(x-5)

得:x=6

3.设工程为a,规定日期为x天,则A每天完成a/x,B每天完成a/(x+3).

[a/x+a/(x+3)] *2 + [a/(x+3)]*(x-2) =a

(2x+3)*2+x(x-2)=x(x+3)

4x+6+x²-2x=x²+3x

x=6

所以规定日期为6天.

看的明白吗?