cách 2 câu 2:tacó:3y=9−x2pt thành:y4+8xy2+16y2−12(y2+4x)+11=0⇔ {x2+4x=1y2+4x=11
TH1:y2+4x=11⇔(9−x23)2+4x=11⇔x4−18x2+36x−18=0
⇔x4=18(x−1)2⇔{x2−3√2x+3√2x=0x2+3√2x−3√2x=0⇔{x=3√2x+hoặc−√18−12√2⇒y=.......2tươngtự=
TH2:y2+4x=1⇔(9−x23)2+4x=1⇔x4−18x2+36x+72=0⇔(x2−6x+12)(x2+6x+6)⇔−3+hoặc−⇒y=−1+hoạc−2√3