Question

分式的乘除计算，要完整的过程 （1）a+b/ab-b²÷[-ab+b²/﹙a-b﹚²]

(2) (1-x)²/x(1-x²).xy+x²y/x-x² (3) x²+xy/x²-xy÷(x+y)÷xy/y²-xy (4) a²-1/a²-4. a+2/a²-2a+1 .a²-4a+4/a+1(5)3a/4y . 2y²/3a² (6) a+2/a-2 . 1/a²+2a (7) 3xy²÷6y²/x

Answer 1

（1）a+b/ab-b²÷[-ab+b²/﹙a-b﹚²]
=(a+b)/b(a-b) . (a-b)²/b(b-a)
=-(a+b)/b²

(2) (1-x)²/x(1-x²).xy+x²y/x-x²
=1/x . x(x+y)/x(1-x)
=(x+y)/(x-x²)

(3) x²+xy/x²-xy÷(x+y)÷xy/y²-xy
=x(x+y)/x(x-y) . 1/(x+y) . y(y-x)/xy
=-1/x

(4) a²-1/a²-4. a+2/a²-2a+1 .a²-4a+4/a+1
=(a+1)(a-1)/(a+2)(a-2) . (a+2)/(a-1)² .(a-2)²/(a+1)
=(a-2)/(a-1)

(5)3a/4y . 2y²/3a²
=(3a×2y×y)/(2×2y×3a×a)
=y/2a

(6) a+2/a-2 . 1/a²+2a
=(a+2)/(a-2).a.(a+2)
=1/(a²-2a)

(7) 3xy²÷6y²/x
=3xy² .x/6y²
=x²/2