J+W 寫到:提供另一種方法:強迫因式分解
∴x+y=6
我最初也想到x+y+xy變成x+y(x+1)，然後我唔知如何計下去.....
因為我已將[1]減了，RHS是14，LHS是x+y(x+1)...........
無奈
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由 lcflcflcf 於 星期三 十二月 22, 2004 11:46 am
我最初也想到x+y+xy變成x+y(x+1)，然後我唔知如何計下去.....
因為我已將[1]減了，RHS是14，LHS是x+y(x+1)...........
無奈
由 J+W 於 星期三 十二月 22, 2004 11:28 am
提供另一種方法:強迫因式分解
∴x+y=6
∴x+y=6
由 lcflcflcf 於 星期三 十二月 22, 2004 10:58 am
1+x+y+xy=15
x+y+xy=14
First, let x be the bigger one
if
y=1, the max num of x is 6
y=2, the max num of x is 4
y cannot be 3, 4, 5 and so on.
So
y=1 or 2
if y=1, x=6,1+6+1*6=13
y cannot be 1
So
y=2
x+2+2x=14
x+2x=12
3x=12
x=4
so x=4, y=2
This is my notion, i sure that there are some solutions which can be firster.
x+y+xy=14
First, let x be the bigger one
if
y=1, the max num of x is 6
y=2, the max num of x is 4
y cannot be 3, 4, 5 and so on.
So
y=1 or 2
if y=1, x=6,1+6+1*6=13
y cannot be 1
So
y=2
x+2+2x=14
x+2x=12
3x=12
x=4
so x=4, y=2
This is my notion, i sure that there are some solutions which can be firster.
[數學]這樣的方程式能解嗎
由 mimi 於 星期三 十二月 22, 2004 10:41 am
請教高手, 這題怎麼解呢?
If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?
拜託! 拜託 !
thanks......
If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?
拜託! 拜託 !
thanks......