y=x^3
x+y=10
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2012-09-13 10:53 am
回答 (3)
2012-09-13 3:57 pm
✔ 最佳答案
你想點解決問題﹖如果xy係整數，y=x^3就有無限個可能，x+y=10亦都有十幾廿個可能，你唔畫圖，唔通你想全部寫哂出嚟﹖
2012-09-13 10:32 pm
Answer: x = 2, y = 8
Solution:
y = x^3 ------------------------------------- ( i )
x + y = 10 --------------------------------- ( ii )
( ii ) - ( i ) => x = 10 - x^3
x^3 + x - 10 = 0 ------------------------- ( iii )
Let f(x) be x^3 + x - 10
f(2) = (2)^3 + (2) - 10 = 0
so x - 2 is a factor of f(x)
x^3 + x - 10 = (x - 2)(ax^2 + bx + c)
x^3 + x - 10 = ax^3 + (b - 2a)x^2 + (c - 2b)x - 2c
we have a = 1, b = 2, c = 5
from ( iii ), we have (x - 2)(x^2 + 2x + 5) = 0
x - 2 = 0 or x^2 + 2x + 5 = 0
x = 2 since (2)^2 - 4(1)(5) = -16 < 0
therefore, x = 2, y = 8
Solution:
y = x^3 ------------------------------------- ( i )
x + y = 10 --------------------------------- ( ii )
( ii ) - ( i ) => x = 10 - x^3
x^3 + x - 10 = 0 ------------------------- ( iii )
Let f(x) be x^3 + x - 10
f(2) = (2)^3 + (2) - 10 = 0
so x - 2 is a factor of f(x)
x^3 + x - 10 = (x - 2)(ax^2 + bx + c)
x^3 + x - 10 = ax^3 + (b - 2a)x^2 + (c - 2b)x - 2c
we have a = 1, b = 2, c = 5
from ( iii ), we have (x - 2)(x^2 + 2x + 5) = 0
x - 2 = 0 or x^2 + 2x + 5 = 0
x = 2 since (2)^2 - 4(1)(5) = -16 < 0
therefore, x = 2, y = 8
參考: knowledge
2012-09-13 4:43 pm
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quisiera ayudarte pero no hablo chino
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