if x+y=4 & xy=15/4
find x^2+y^2
help/....
maths........
2009-05-17 1:13 am
回答 (3)
2009-05-17 1:22 am
✔ 最佳答案
x+y=4(x+y)2=42
x2+y2 -2xy=16
x2+y2 -2(15/4)=16
x2+y2 =47/2
2009-05-16 17:23:53 補充：
(x+y)^2=4^2
x^2+y^2 +2xy=16
x^2+y^2 +2(15/4)=16
x^2+y^2 =17/2
2009-05-17 3:47 am
x+y=4 ---- (1)
xy=15/4 ----(2)
(1)^2,
(x+y)^2=4^2
x^2+2xy+y^2=16----(3)
Sub (2)into (3)
x^2+y^2+2xy=16
x^2+y^2+ (15/4)(2)=16
x^2+y^2=16-15/2
x^2+y^2=32/2-15/2
x^2+y^2=17/2
xy=15/4 ----(2)
(1)^2,
(x+y)^2=4^2
x^2+2xy+y^2=16----(3)
Sub (2)into (3)
x^2+y^2+2xy=16
x^2+y^2+ (15/4)(2)=16
x^2+y^2=16-15/2
x^2+y^2=32/2-15/2
x^2+y^2=17/2
參考: me
2009-05-17 1:50 am
x + y = 4
(x+y)2 = 4^2
x^2 + y^2 + 2xy = 16
x^2 + y^2 + 2(15/4) = 16
x^2 + y^2 = 16 - 15/2
x^2 + y^2 = 17/2
希望幫到你！！
(x+y)2 = 4^2
x^2 + y^2 + 2xy = 16
x^2 + y^2 + 2(15/4) = 16
x^2 + y^2 = 16 - 15/2
x^2 + y^2 = 17/2
希望幫到你！！
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