解1题
原式=[(x+y)/x]-[x/(x-y)]+[y²/(x²-xy)]
=[(x+y)/x]-[x/(x-y)]+{y²/[x(x-y)]}
={(x+y)(x-y)/[x(x-y)]}-{x²/[x(x-y)]}+{y²/[x(x-y)]}
=[(x+y)(x-y)-x²+y²]/[x(x-y)]
=(x²-y²-x²+y²)/[x(x-y)]
=0
解2题
原式=[x/(x-3)]-[(x+6)/(x²-3x)]+1/x
=[x/(x-3)]-{(x+6)/[x(x-3)]+1/x
={x²/[x(x-3)]}-{(x+6)/[x(x-3)]}+{(x-3)/[x(x-3)]}
=[x²-(x+6)+(x-3)]/[x(x-3)]
=(x²-x-6+x-3)/[x(x-3)]
=(x²-9)/[x(x-3)]
=(x+3)(x-3)/[x(x-3)]
=(x+3)/x
=(200+3)/200
=203/200