已知x+y=3,xy=-1,求x四次方+y四次方的值

网友回答

x^2+y^2=(x+y)^2-2xy=3^2-2*(-1)=9+2=11
x^4+y^4=(x^2+y^2)^2-2x^2y^2=11^2-2*(-1)^2=121-2=119
======以下答案可供参考======
供参考答案1:
∵x^2+y^2=(x+y)^2-2xy=9+2=11,
∴x^4+y^4=[((x+y)^2-2xy)]^2-2(xy)^2
=11×11-2×1=119.
供参考答案2:
我[email protected]用表示m的n次方 [email protected][email protected]=([email protected][email protected])@2-2*[email protected]*[email protected]=((x+y)@2-2xy)@2-2*[email protected]*[email protected]=119