已知多项式kx2-6xy-8y2可分解为2,（mx+y)(x-4y),求k,m的值.若x2+y2+4x-6y+13=0,x,y均为有理数,则xy是多

Question

已知多项式kx2-6xy-8y2可分解为2,（mx+y)(x-4y),求k,m的值.若x2+y2+4x-6y+13=0,x,y均为有理数,则xy是多

Answer 1

你好,蓝love小海
1、
2(mx+y)(x-4y)
=2(mx²-4mxy+xy-4y²)
=2[mx²+(1-4m)xy-4y²]
=2mx²+2(1-4m)xy-8y²
=kx²-6xy-8y²
∴k=2m
2(1-4m)=-6
解得m=1,k=2
2、
∵x²+y²+4x-6y+13=0
x²+4x+4+y²-6y+9=0
(x+2)²+(y-3)²=0
且(x+2)²≥0,(y-3)²≥0
∴x+2=0,y-3=0
解得x=-2,y=3
∴xy=(-2)×3=-6