1.若a,b,x,y满足ax+by=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax^5+by^5的值2.已知等式k²x+2(k-1)y+(2-k-k²)z=1与k值无关,求x,y,z的值

Question

1.若a,b,x,y满足ax+by=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax^5+by^5的值
2.已知等式k²x+2(k-1)y+(2-k-k²)z=1与k值无关,求x,y,z的值

Answer 1

1.若a,b,x,y满足ax+by=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax^5+by^5的值
ax+by=3
ax^2+by^2=7
所以
a=(7-3y)/(x^2-xy)
b=(7-3x)/(y^2-xy)
ax^3+by^3=x^2(7-3y)/(x-y)+y^2(7-3x)/(y-x)=(7x^2-3x^2y-7y^2+3xy^2)/(x-y)
=[7(x+y)(x-y)-3xy(x-y)]/(x-y)
=7x+7y-3xy=16
3xy=7(x+y)-16
ax^4+by^4=x^3(7-3y)/(x-y)+y^3(7-3x)/(y-x)=(7x^3-3x^3y-7y^3+3xy^3)/(x-y)
=[7(x-y)(x^2+xy+y^2)-3xy(x-y)(x+y)]/(x-y)
=7x^2+7xy+7y^2-[7(x+y)-16](x+y)
=7x^2+7xy+7y^2-7x^2-14xy-7y^2+16x+16y
=-7xy+16x+16y
=42
所以
7(x+y)-3xy=16
16(x+y)-7xy=42
所以x+y=-14
xy=-38
x^2+y^2=(x+y)^2-2xy=272
ax^5+by^5=x^5*(7-3y)/(x^2-xy)+y^5(7-3x)/(y^2-xy)
=(7x^4-3x^4y-7y^4+3xy^4)/(x-y)
=7(x+y)(x^2+y^2)-3xy(x^2-xy+y^2)
把x+y=-14
xy=-38
x^2+y^2=272代入
=8684
2.已知等式k²x+2(k-1)y+(2-k-k²)z=1与k值无关,求x,y,z的值
k^2x+2ky-2y+2z-kz-k^2z=1
(x-z)k^2+(2y-z)k+(-2y+2z-1)=0*k^2+0*k+0
所以x-z=0
2y-z=0
-2y+2z-1=0
x=1,y=1/2,z=1