數學知識交流 － 求值 (2)

x + y = 3
==>
x² + y² = 9 - 2xy ......(1) x⁴+ y⁴= (x² + y²)² - 2x²y² = 24 ......(2)
代 (1) 入 (2) :(9 - 2xy)² - 2x²y² = 24
2(xy)² - 36xy + 57 = 0
xy = 9 ± √210 /2
x(3 - x) = 9 ± √210 /2
x² - 3x + 9 ± √210 /2 = 0

因 x 為實數 , 只有 x² - 3x + 9 - √210 /2 = 0 , 故 xy = 9 - √210 /2 ,
(x - y)²
= x² + y² - 2xy
= 9 - 4xy ..... (由(1))
= 9 - 4(9 - √210 /2)
= 2√210 - 27x - y = ±√ (2√210 - 27) ***************************************************************************************

1)x³ + y³
= (x + y) (x² - xy + y²)
= (3) (9 - 3xy) ..... (由(1))
= 27 - 9 (9 - √210 /2)
= 9√210 /2 - 54 所有可能值之和 = 9√210 /2 - 54
2)x² - y²
= (x - y) (x + y)
= ±√ (2√210 - 27) (3)所有可能值之和 = 0
3)x⁵- y⁵
= (x - y) ( x⁴+ x³y + x²y² + xy³ + y⁴)
= (x - y) ( x⁴+ y⁴+ xy(x² + xy + y²) )
= (x - y) ( x⁴+ y⁴+ xy ((x + y)² - xy) )
= ±√ (2√210 - 27) (24 + (9 - √210 /2) (3² - (9 - √210 /2)) ) 所有可能值之和 = 0
註 :
x = (1/2) (3 - √ (2√210 - 27) ) 或 x = (1/2) (3 + √ (2√210 - 27) )
y = (1/2) (3 + √ (2√210 - 27) ) 或 y = (1/2) (3 - √ (2√210 - 27) )
其實 2) 和 3) 題不用計也知是 0 。