Tìm $m$ để hệ có nghiệm duy nhất: 
      $y^2=x^3-4x^2+mx$
      $x^2=y^3-4y^2+my$  
$lấy (1)-(2) => (x-y)[x^2+xy+y^2-3(x+y)+m]=0$ (*)
$x=y$ vì pt có nghiệm duy nhất 
=> (*)=>$-6x+3x^2+m=0$
$\Delta '=(-3)^2-3.m=0$
=> $m=3$
em làm theo pp đk cần và đủ anh ak –  Kiyoshi Bụt 12-12-16 06:38 AM
e tính denta thì sao mà vô nghiệm đc –  Ryo 12-12-16 06:36 AM
ko phải anh ak ý em là nếu th sau la vô ng thì sao –  Kiyoshi Bụt 12-12-16 06:30 AM
k trình bày đấy –  Ryo 12-12-16 06:25 AM
có cái j thiếu thiếu thì phải –  Kiyoshi Bụt 12-12-16 06:24 AM
me mà nị –  ๖ۣۜTQT☾♋☽ 10-12-16 05:17 AM
v tùng e. tn dùng anh lộn việt hay z –  Effort 10-12-16 04:50 AM
cái nick old of you đâu mà now dùng nick this –  ๖ۣۜTQT☾♋☽ 10-12-16 01:09 AM
thế x=y s ko lây x-y=0 mak láy cái sau =0 hak ca –  Effort 09-12-16 08:09 AM

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