x+y=10
平方得
x^2+y^2+2xy=100
xy=50-(x^2+y^2)/2
x³+y³=100
=(x+y)(x^2+y^2-xy)
=10*[x^2+y^2-50+(x^2+y^2)/2]
3(x^2+y^2)/2=60
x^2+y^2=40
(x+y)3=x3+y3+3xy(x+y)
得xy=30
(x+y)2=x2+2xy+y2所以
x2+y2=(x+y)2-2xy=100-2x30=40
嘿嘿不上学好多年~~~~
x³+y³=(x+y)(x²-xy+y²)=100 , x²+y²=10+xy ,xy=x²+y²-10
x+y=10 , x²+y²+2xy=100 ,x²+y²=100-2xy=100-2(x²+y²-10)
3(x²+y²)=120
x²+y²=40
解:∵x³ +y³ = (x+y)【(x+y)² - 3xy】 = 100
∴ 10 (100 - 3xy) = 100 xy = 30
∴x²+y² = (x+y)² - 2xy = 100 - 60 = 40
(x+y)³=x³+3x²y+3xy²+y³ x³+y³=(x+y)(x^2-xy+y^2)=1000
x²-xy+y²=10 ①
﹙x+y﹚²=x²+y²+2xy=100 ②
①x2+②得出 3(x²+y²)=120
x²+y²=40