Question

x y∈R 且x+y＝3 1/(x+y²)+1/(x²+y)＝1/2 求x∧5+y∧5，

谁真的答案哦？

Answer 1

绝对对称式求值，先求 xy
2(x^2+y^2+x+y)=x^2y^2+x^3+y^3+xy
2[(x+y)^2-2xy+x+y]=(xy)^2+(x+y)[(x+y)^2-3xy]+xy
2[12-2xy]=(xy)^2+3[9-3xy]+xy
(xy)^2-4xy+3=0
xy=1 or xy=3
t^2-3t+3=0，△=-3<0
xy=3 舍去，所以 xy=1
x^5+y^5=(x^2+y^2)(x^3+y^3)-(xy)^2(x+y)
=[(x+y)^2-2xy](x+y)[(x+y)^2-3xy]-(xy)^2
=7*3*6-1=125

Answer 2

123
解析：
2●[1/(x+y²)+1/(x²+y)]=1
2(x²+y+x+y²)=(x+y²)(x²+y)
2[(x+y)+(x²+y²)]=xy+x²y²+x³+y³
2[3+(x+y)²-2xy]=xy+x²y²+(x+y)[(x+y)²-3xy]
2(12-2xy)=(xy)²+xy+3(9-3xy)
(xy)²-4xy+3=0
xy=1或3
检验，
xy=3时，|x+y|=|x+3/x|≥2√3，与x+y=3矛盾。
故，xy=3舍去。
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x²+y²
=(x+y)²-2xy
=3²-2●1
=7
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x³+y³
=(x+y)[(x+y)²-3xy]
=3●(3²-3)
=18
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x^5+y^5
=(x²+y²)(x³+y³)-(x²y³+x³y²)
=7●18-(xy)²●(x+y)
=126-3
=123

Answer 3

供参考

Answer 4

∠ABC=180-40-20-50=70度

Answer 5

设u=x+y=3,v=xy,则
1/(x+y²)+1/(x²+y)
=（x+y^2+x^2+y）/(x+y^2)(x^2+y)
=(u+u^2-2v)/[u(u^2-3v)+v+v^2)
=(3+9-2v)/[3(9-3v)+v+v^2]＝1/2,
所以24-4v=27-9v+v^2,
所以v^2-5v+3=0,
所以v=(5-√13)/2,（舍弃(5+√13)/2）
所以x^5+y^5
=u^5-5u^3v+5uv^2
=243-135v+15v^2
=243-135v+15(5v-3)
=198-60v
=198-30(5-√13)
=48+30√13.