Question

已知方程组x+ky=3;2x²+y²=6(k>1)的两组实数解为（x1,y1)和（x2,y2),且y

Answer 1

①得,x=3-ky③

代入②得,18-12ky+2k²y²+y²=6

(2k²+1)y²-12ky+12=0

∵两组实数解为（x1,y1)和（x2,y2),

∴y1+y2=12/(2k²+1) x1+x2=6-k(y1+y2)=6-12k/(2k²+1)=(12k²-12k+6)/(2k²+1)

∵y1+y2=4(x1+x2),

∴12/(2k²+1)=4(12k²-12k+6)/(2k²+1)

∴2k²-2k=0

∴k1=0 k2=1

∵⊿=(-12k)²-4(2k²+1)≥0

∴k=1