Question

已知:x+y=3,x^2+y^2+xy=8,则x^4+y^4+3x^4y+3xy^4的值?

Answer 1

x+y=3
两边平方得x^2+y^2+2xy=9
又x^2+y^2+xy=8
两式相减得xy=1
代入得x^2+y^2=7
两边平方得x^4+y^4+2x^2y^2=49
所以x^4+y^4=47
x^4+y^4+3x^4y+3xy^4
=（x^4+y^4）+3xy(x^3+3xy^3)
=47+3(x+y)(x^2-xy+y^2)
=47+3×3（7-1）
=101

Answer 2

(x+y)²-(x²+y²+xy)
=(x²+2xy+y²)-(x²+y²+xy)
=xy=3²-8=1

x+y=3 xy=1

x^4+y^4+3x^4y+3xy^4
=x^4+y^4+3xy(x^3+y^3)
=(x²+y²)²-2x²y² +3xy(x+y)(x²-xy+y²)
=[(x+y)²-2xy]²-2x²y²+3xy(x+y)(x²+y²+xy-2xy)
=(3²-2)²-2×1²+3×1×3×(8-2×1)
=101

追问

=(x²+y²)²-2x²y² +3xy(x+y)(x²-xy+y²)不明白为什么不是=(x²+y²)²-2x²y² +3xy(x+y)(x²-2xy+y²)呢？

Answer 3

x^2+y^2+xy=8
x^2+y^2+2xy-xy=8
(x+y)^2-xy=8
3^2-xy=8
9-xy=8
xy=1

x^2+y^2+xy=8
x^2+y^2+1=8
x^2+y^2=7

x^4+y^4+3x^4y+3xy^4
=x^4+y^4+2x^2y^2-2x^2y^2+3x^4y+3xy^4
=(x^2+y^2)^2-2(xy)^2+3xy(x^3+y^3)
=7^2-2*1^2+3(x^3+y^3)
=49-2+3(x^3+y^3)
=47+3(x^3+y^3)
=47+3(x+y)(x^2+y^2-xy)
=47+3*3*(7-1)
=47+54
=101

Answer 4

x+y=3两边平方在减去x^2+y^2+xy=8，所以xy=1，所以x^4+y^4+3x^4y+3xy^4xy+3x^3+3y^3)(xy)=(x^3+y^3+3x^3+3y^3)=4(x^3+y^3)=4(x+y)(x^2-xy+y^2)=12(x^2-xy+y^2)=12[(x+y)^2-3xy]=72

追问

错了吧

追答

。。。
没用草稿纸可能错了sorry……

追问

改正啊！

追答

LS其实说对了，立方和公式就是这样的

追问

你的答案真的不对！

追答

我是说我上面的那个回答者没说我自己答对了啊……………………