[5xy^2(x^2-3xy)-(-3x^2y^3)^3]除(1⼀2xy)^2 2问:已知xy是有理数且x+y=4,x^2+y^2=14,求x^7+y^7的值

2024-12-13 04:57:54
推荐回答(1个)
回答1:

[5xy^2(x^2-3xy)-(-3x^2y^3)^3]除(1/2xy)^2
=[5x²y²(x-3y)+27x^6y^9]÷(1/4*x²y²)
=[5x²y²(x-3y)+x²y²(27x^4y^7)]÷(1/4*x²y²)
=4[5(x-3y)+27x^4y^7]
=20x-60y+98x^4y^7

∵x+y=4,x^2+y^2=14,
∴2xy=(x+y)²-(x²+y²)=16-14=2
∴xy=1
∴(x²+y²)²=196
∴x⁴+y⁴+2x²y²=196
∴x⁴+y⁴=194
x³+y³=(x+y)(x²-xy+y²) =4*(14-1)=52
∴x^7+y^7
=(x³+y³)(x⁴+y⁴)-x³y⁴-x⁴y³
=52*194-x³y³(x+y)
=52*194-1*4
=10084