Решить систему уравнений
|2х + у = 1
|2х^2 + ху + у^2 = 1
и
|3x + 2y = 5
|2x^2 + 3y = 12
|2х + у = 1
|2х^2 + ху + у^2 = 1
и
|3x + 2y = 5
|2x^2 + 3y = 12
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Answers & Comments
| 2x²+x(1-2x)+(1-2x)²=1
| y=1-2x
| 2x²+x-2x²+(1-4x+4x²)=1
| y=1-2x
| x+1-4x+4x²-1=0
| y=1-2x
| 4x²-3x=0
| y=1-2x
| x(4x-3)=0
| y=1-2x или | y=1-2x
| x=0 | 4x-3=0
| y=1-2*0 | y=1-2*3/4
| x=0 | x=3/4
| y= 1 | y=-0,5
| x=0 | x=0,75
Проверка
| 2*0+1=1
|2*0²-0*1+1²=1
| 2*0.75+(-0.5)=1
| 2 *0.75²-0.75*(-0.5)+(-0.5)²=1
Ответ (0;1), (0,75;-0,5)
2) |3x + 2y = 5
|2x² + 3y = 12
|y=(5-3x)/2
|2x² + 3(5-3x)/2=12
|y=(5-3x)/2
| 4x²+3(5-3x)=24
4x²+3(5-3x)=24
4x²+15-9x=24
4x²+15-9x-24=0
4x²-9x-9=0 за т Виетта ищем корени
D=9²-4*4*(-9)=81+144=225
√D=√225=15
x1=4 или x2=-1
|y=(5-3x)/2 или |y=(5-3x)/2
| x=4 | x=-1
|y=(5-3*4)/2 |y=(5-3*(-1))/2
| x=4 | x=-1
| y=-3.5 | y=4
| x=4 | x=-1
Ответ , (4;-3.5), (-1;4)