Question

已知x+y=1,x^2+y^2=2 求x^7+y^7

x加y等于1,x的平方加y的平方等于2 求x的7次方加y的7次方 过程全点

Answer 1

因为x+y=1,x^2+y^2=2, 又因为x^2+y^2=(x+y)^2-2xy=1^2-2xy=2, 所以2xy=-1, 所以xy=-1/2, 所以x^3+y^3=(x+y)(x^2-xy+y^2)=1*(2-xy)=1*[2-(-1/2)]=2.5, x^4+y^4=(x^2+y^2)^2-2x^2y^2=2^2-2*(-1/2)^2=3.5, 所以x^7+y^7=(x^3+y^3)(x^4+y^4)-x^3y^4-x^4y^3 =(x^3+y^3)(x^4+y^4)-(xy)^3(x+y) =2.5*3.5-(-0.5)^3*1 =8.75-(-0.125) =8.875.