Question

若y/x+x/y=-2,求代数式2x^2+3xy-2y^2/x^2+xy-2y^2的值

Answer 1

y/x+x/y=-2
设Y/X=A
则A+1/A=-2 ==>A^2+2A+1=0==>(A+1)^2=0 ==>A=-1
所以(2x^2+3xy-2y^2)/(x^2+xy-2y^2)=(2X-Y)(X+2Y)/((X+2Y)(X-Y))
=(2X-Y)/(X-Y)=X/(X-Y)+1=1/(1-Y/X)+1=1/(1-(-1))+1=3/2

Answer 2

y/x+x/y=-2
(x^2+y^2)/xy=-2
x^2+2xy+y^2=0
(x+y)^2=0
x+y=0即x=-y
2x^2+3xy-2y^2/x^2+xy-2y^2
=2x^2-3x^2-2x^2/x^2-x^2-2x^2
=x^2/-2x^2
=-1/2

Answer 3

y/x+x/y=-2
y^2+x^2=-2xy
x=-y
带入代数式2x^2+3xy-2y^2/x^2+xy-2y^2
则变成2y^-3y^2-2y^2/y^2-y^2-2y^2
=-3y^2/-2y^2
=3/2

Answer 4

1.5