a4次+b4次 點計

a^4 + b^4
=(a² )² + (b² )²
=(a² )² + 2(a² )(b² ) + (b² )²-2(a² )(b² )
=(a²+ b²)² - 2(ab)²
=(a² + 2ab + b² - 2ab)² - 2(ab)²
=( (a+b)² - 2ab )²-2(ab)²
=( (-2)² - 2(7/3) )² - 2(7/3)²
=25/9 - 98/9
=-73/9

2006-10-27 22:02:20 補充：
正確ｄ應該係a^4 b^4=(a² )² (b² )² =(a² )² 2(a² )(b² ) (b² )²-2(a² )(b² )=(a² b²)² - 2(ab)²=(a² 2ab b² - 2ab)² - 2(ab)²=( (a b)² - 2ab )²-2(ab)²=( (-2)² - 2(7/3) )² - 2(7/3)²=4/9 - 98/9=-94/9

a^4+b^4
=(a^2)^2+(b^2)^2
=(a^2)^2+(b^2)^2+2(a^2)(b^2)-2(a^2)(b^2)
=[(a^2)+(b^2)]^2-2(a^2)(b^2)
=[a^2+b^2+2ab-2ab]^2-2(ab)^2
=[(a+b)^2-2ab]^2-2(ab)^2
=[(-2)^2-2(7/3)]^2-2(7/3)^2
=[4-14/3]^2-2(49/9)
=(-2/3)^2-98/9
=4/9-98/9
=-94/9

a^4 + b^4
=[a^4+2(a^2)(b^2)+b^4]-2(a^2)(b^2)
=(a^2+b^2)^2-2(a^2)(b^2)
=[(a^2+2ab+b^2)-2ab]^2-2(ab)^2
=[(a+b)^2-2ab]^2-2(ab)^2
=[(-2)^2-2x(7/3)]^2-2x(7/3)^2
=[4-(14/3)]^2-(98/9)
=(-2/3)^2-(98/9)
=-94/9

a^4+b^4
=(a+b)^4-4a^3b-6a^2b^2-4ab^3
=(a+b)^4-2ab(2a^2+3ab+2b^2)
=(a+b)^4-2ab(3ab+2(a^2+b^2))
=(a+b)^4-2ab(3ab+2((a+b)^2-2ab))
=(-2)^4-2(7/3)(3(7/3)+2((-2)^2-2(7/3)))
=16-14/3(7+2(4-14/3))
=16-14/3(7+8-28/3)
=16-14/3(17/3)
=16-238/3
=-190/3

a^4+b^4
=(a+b)^4-4a^3b-6a^2b^2-4ab^3
=(a+b)^4-2ab(2a^2+3ab+2b^2)
=(a+b)^4-2ab(3ab+2(a^2+b^2))
=(a+b)^4-2ab(3ab+2((a+b)^2-2ab))
=(-2)^4-2(7/3)(3(7/3)+2((-2)^2-2(7/3)))
=16-14/3(7+2(4-14/3))
=16-14/3(7+8-28/3)
=16-14/3(17/3)
=16-238/3
=-190/3