решите систему уравнений методом подстановки а)y=2-x x2+4y=8 Б)x-y=4 X2+xy=6

Ответы 2

а)y=2-x x2+4y=8x²+4(2-x)-8=0x²-4x=0x(x-4)=0x=0⇒y=2x=4⇒y=-2(0;2);(4;-2)Б)x-y=4⇒y=x-4X2+xy=6x²+x(x-4)-6=0x²+x²-4x-6=02x²-4x-6=0x²-2x-3=0x1+x2=2 U x1*x2=-3x1=-1⇒y=-5x2=3⇒y=-1(-1;-5);(3;-1)
- Автор:
  alma3odx
- 5 лет назад
- 0
$\left \{ {{y=2-x} \atop {x^2+4y=8}} ight. \\ \\ \left \{ {{y=2-x} \atop {x^2+4(2-x)=8}} ight. \\ \\ \left \{ {{y=2-x} \atop {x^2+8-4x-8=0}} ight. \\ \\ \left \{ {{y=2-x} \atop {x^2-4x=0}} ight. \\ \\ \left \{ {{y=2-x} \atop {x(x-4)}} ight. \\ \\ \left \{ {{x=0} \atop {y=2}} ight. i li \left \{ {{x=4} \atop {y=-2}} ight. \\ \\$ $\left \{ {{x-y=4} \atop {x^2+xy=6}} ight. \\ \\ \left \{ {{x=y+4} \atop {(y+4)^2+(y+4)y=6}} ight. \\ \\ \left \{ {{x=y+4} \atop {y^2+8y+16+y^2+4y-6=0}} ight. \\ \\ \left \{ {{x=y+4} \atop {2y^2+12y+10=0}} ight. \\ \\ \left \{ {{x=y+4} \atop {2(y^2+6y+5)=0}} ight. \\ \\ \left \{ {{x=y+4} \atop {y^2+6y+5=0}} ight. \\ \\ \left \{ {{x=y+4} \atop {y^2+5y+y+5=0}} ight. \\ \\ \left \{ {{x=y+4} \atop {(y+5)(y+1))=0}} ight. \\ \\$ $\left \{ {{x=-1} \atop {y=-5}} ight. ili \left \{ {{x=3} \atop {y=-1}} ight. \\ \\$
- Автор:
  antony
- 5 лет назад
- 0
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