Question

Помогите решить пожалуйста

Answer 1

$\left \{ {{x^{2}+xy+\frac{1}{4}y^{2}-x-\frac{1}{2}y=2} \atop {\frac{1}{4}x^{2}-xy+y^{2}+2y-x=3}} \right. \\\\\left \{ {{(x+\frac{1}{2}y)^{2}-(x+\frac{1}{2}y)=2} \atop {(\frac{1}{2}x-y)^{2}-2(\frac{1}{2} x-y)=3}} \right. \\\\x+\frac{1}{2}y=a;\frac{1}{2}x-y=b\\\\\left \{ {{a^{2}-a-2=0 } \atop {b^{2}-2b-3=0}} \right. \\\\\left \{ {{\left[\begin{array}{ccc}a_{1}=2 \\a_{2}=-1 \end{array}\right } \atop {\left[\begin{array}{ccc}b_{1}=3 \\b_{2}=-1 \end{array}\right }} \right.$

$1)\left \{ {{a=2} \atop {b=3}} \right.\\\\ \left \{ {{x+\frac{1}{2}y=2}|*2 \atop {\frac{1}{2}x-y=3 }} \right.\\\\+\left \{ {{2x+y=4} \atop {\frac{1}{2}x-y=3}} \right.\\ ------\\2,5x=7\\\\x=2,8\\\\y=\frac{1}{2}x-3=\frac{1}{2}*2,8-3=-1,6\\\\2)\left \{ {{a=2} \atop {b=-1}} \right.\\\\\left \{ {{x+\frac{1}{2}y=2}|*2 \atop {\frac{1}{2}x-y=-1 }} \right.\\\\+\left \{ {{2x+y=4} \atop {\frac{1}{2}x-y=-1 }} \right.\\------\\ 2,5x=3\\\\x=1,2\\\\y=1+\frac{1}{2}x=1+\frac{1}{2}*1,2=1+0,6=1,6$

$3)\left \{ {{a=-1} \atop {b=3}} \right.\\\\\left \{ {{x+\frac{1}{2}y=-1}|*2 \atop {\frac{1}{2}x-y=3 }} \right.\\\\+\left \{ {{2x+y=-2} \atop {\frac{1}{2}x-y=3 }} \right. \\------\\2,5x=1\\\\x=0,4\\\\y=\frac{1}{2}x-3=\frac{1}{2}*0,4-3=-2,8\\\\4)\left \{ {{a=-1} \atop {b=-1}} \right. \\\\\left \{ {{x+\frac{1}{2}y=-1}|*2 \atop {\frac{1}{2}x-y=-1}} \right.\\\\+\left \{ {{2x+y=-2} \atop {\frac{1}{2}x-y=-1}} \right.\\------\\ 2,5x=-3\\\\x=-1,2$

$y=\frac{1}{2}x+1=\frac{1}{2}*(-1,2)+1=-0,6+1=0,4\\\\Otvet:\boxed{(2,8;-1,6),(1,2;1,6),(0,4;-2,8),(-1,2;0,4)}$

Answer 2

Ответ:

Объяснение:

Система уравнений:

x²+xy +1/4 ·y²-x -1/2 ·y=2; (x +1/2 ·y)²-(x +1/2 ·y)=2

1/4 ·x²-xy+y²+2y-x=3; (1/2 ·x-y)²+2y-x=3   |2

(x+1/2 ·y)²-(x +1/2y)-2=0

1/2 ·(1/2 ·x-y)²-(1/2 ·x-y)=3/2; 1/2 ·(1/2 ·x-y)²-(1/2 ·x-y) -1,5=0

(x +1/2 ·y)²-(x +1/2y)-2=0; t=x +1/2 ·y

t²-t-2=0

t₁+t₂=1; -1+2=1

t₁·t₂=-2; -1·2=-2

t₁=-1; t₂=2

x +1/2 ·y=-1; x +1/2 ·y=2

1/2 ·(1/2 ·x-y)²-(1/2 ·x-y) -1,5=0; r=1/2 ·x-y

1/2 ·r²-r-1,5=0       |×2

r²-2r-3=0

r₁+r₂=2; -1+3=2

r₁·r₂=-3; -1·3=-3

r₁=-1; r₂=3

1/2 ·x-y=-1; 1/2 ·x-y=3

Составляем первую систему уравнений:

x +1/2 ·y=-1

1/2 ·x-y=-1

x +1/2 ·y=1/2 ·x-y

1/2 ·x-x=1/2 ·y+y

-1/2 ·x=3/2 ·y

x=3/2 ·y·(-2); x=-3y

-3y +1/2 ·y=-1; 5/2 ·y=1; y=2/5=0,4

x=-3·2/5=-6/5=-1,2

И вторую систему уравнений:

x +1/2 ·y=2

1/2 ·x-y=3

1/2 ·x-y-x -1/2 ·y=3-2

-1/2 ·x -3/2 ·y=1

1/2 ·x=-3/2 ·y-1

x=2(-3/2 ·y-1); x=-3y-2

-3y-2 +1/2 ·y=2

-5/2 ·y=2+2

y=4·(-2/5)=-8/5=-0,6

x=-3·(-8/5) -2=4,8-2=2,8

Затем третью систему уравнений:

x +1/2 ·y=-1

1/2 ·x-y=3; y=1/2 ·x-3

x +1/2 ·(1/2 ·x-3)=-1

x +1/4 ·x=-1 +3/2

5/4 ·x=1/2; x=1/2 ·4/5=2/5=0,4

y=1/2 ·2/5 -3=1/5 -3=-2,8

И четвёртую систему уравнений:

x +1/2 ·y=2

1/2 ·x-y=-1

x +1/2 ·y +1/2 ·x -y=2-1

3/2 ·x -1/2 ·y=1

1/2 ·y=3/2 ·x-1

y=2(3/2 ·x-1); y=3x-2

1/2 ·x-3x+2=-1

5/2 ·x=1+2

x=3·2/5=6/5=1,2

y=3·6/5 -2=18/5 -2=3,6-2=1,6

Итого получаем корни: (-1,2; 0,4); (2,8; -0,6); (0,4; -2,8); (1,2; 1,6).