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09 Hope Cup Mathematics The 20th World Second Attempt to Solve the First Problem 09 Hope Cup Mathematics The 20th World Second Attempt to Solve the penultimate problem. ...
It is said that there are many people asking this question today.

Then I Ctrl+V a paragraph from my notebook.

Below.

According to the meaning of the question

2x+xy+x/y= 100

2xy+xyy+x= 100y

(yy+2y+ 1)x= 100y

(y+ 1)(y+ 1)x = 100y

100=2*2*5*5

So it is divided into two identical factors:

100= 1* 1

*

100

100=2*2

*

25

100=5*5

*

four

100= 10* 10

*

1

Let (y+ 1)(y+ 1) be equal to 1 * 1, 2 * 2, 5 * 5, 10 * 10 respectively.

Y+ 1= plus or minus 1, plus or minus 2, plus or minus 5, plus or minus 10.

The solution is y=0, -2, 1, -3, 4, -6, 9,-1 1, where y = 0 does not meet the meaning of the question.

Then replace the original equation operation with these y values.

The final answer is short.

(I said I did something wrong-

-)