初中数学题拜托诸位!(其三)[有难度]

y/x(x+y)=1/x-1/(x+y)
通分即可证明1/x-1/(x+y)=(x+y-x)/x(x+y)=y/x(x+y)
y/x(x+y)+y/(x+y)(x+2y).....+y/(x+9y)(x+10y)
=1/x-1/(x+y)+1/(x+y)-1/(x+2y)....+1/(x+9y)-1/(x+10y)
=1/x-1/(x+10y)
=(x+10y-x)/x(x+10y)
=10y/x(x+10y)

真受不了，这也叫难度

y/x(x+y)
=(y+x-x)/x(x+y)
=(x+y)/x(x+y)-x/x(x+y)
=1/x-1/(x+y)

y/(x+y)(x+2y)
=(2y+x-x-y)/(x+y)(x+2y)
=(x+2y)/(x+y)(x+2y)-(x+y)/(x+y)(x+2y)
=1/(x+y)-1/(x+2y)

同理y/(x+9y)(x+10y)=1/(x+9y)-1/(x+10y)

所以原式=[1/x-1/(x+y)]+[1/(x+y)-1/(x+2y)]+……+[1/(x+9y)-1/(x+10y)]
=1/x-1/(x+10y)
=(x+10y-x)/x(x+10y)
=10y/x(x+10y)

y/x(x+y)=1/x-1/(x+y)=(x+y)/x(x+y)-y/x(x+y)
所以初中数学题拜托诸位!(其三)[有难度]

y[1/x(x+y)+1/(x+y)(x+2y)+...+1/(x+9y)(x+10y)
=y[1/x-1/(x+y)+1/(x+y)-1/(x+2y)+.....1/(x+9y)-1/(x+10y)]
=y=[1/x-1/(x+10y)]

这题是用这公式,1/x-1/(x+1)=1/(x+1)x

此题就如你说的做法
y提出来没问题
只需说其中一个，1/x(x+y)=1/y【1/x-1/(x+y)】这可以说已经是公式了
换个例子1/a(a