Question

с заданиями их не так много
Первые 3 метод подстановки

Answer 1

1.

$\left \{ {{2x-y=7} \atop {x^2-2y=19}} \right. \left \{ {{-y=7-2x} \atop {x^2-2y=19}} \right. \left \{ {{y=2x-7} \atop {x^2-2\cdot (2x-7)=19}} \right. \left \{ {{y=2x-7} \atop {x^2-4x+14=19}} \right. \left \{ {{y=2x-7} \atop {x^2-4x+14-19=0}} \right. \left \{ {{y=2x-7} \atop {x^2-4x-5=0}} \right. \\ \\ x^2-4x-5=0 \\ \\ x_1+x_2=4 \\ x_1\cdot x_2=-5 \\ \\ x_1 =5; \ \ \ x_2=-1 \\\\ y=2x-7; \\ \\ y_1=2\cdot 5 -7=10-7=3 \\\\ y_2=2\cdot (-1)-7=-2-7=-9$

$\left \{ {{y_1=3; \ \ \ y_2=-9} \atop {x_1=5; \ \ \ x_2=-1}} \right.$

2.

$\left \{ {{x^2+y^2=13} \atop {y-x=-1}} \right. \left \{ {{x^2+(x-1)^2=13} \atop {y=x-1}} \right. \left \{ {{x^2+x^2-2x+1=13} \atop {y=x-1}} \right. \left \{ {{2x^2-2x+1-13=0} \atop {y=x-1}} \right. \left \{ {{2x^2-2x-12=0} \atop {y=x-1}} \right. \\ \\ \left \{ {{x^2-x-6=0} \atop {y=x-1}} \right. \\ \\ x_1+x_2=1 \\ x_1\cdot x_2=-6 \\ \\ x_1=-2; \ \ \ x_2=3 \\ \\ y=x-1 \\ \\ y_1=-2-1=-3; \ \ \ \ y_2 =3-1=2$

$\left \{ {{x_1=-2; \ \ \ x_2 = 3 } \atop {y_1=-3; \ \ \ y_2 =2}} \right.$

3.

$\left \{ {{xy=4} \atop {x+y=5}} \right. \left \{ {{(5-y)\cdot y=4} \atop {x=5-y}} \right. \left \{ {{5y-y^2-4=0} \atop {x=5-y}} \right. \left \{ {{y^2-5y+4=0} \atop {x=5-y}} \right. \\ \\ y^2-5y+4=0 \\ \\ y_1+y_2=5\\ y_1\cdot y_2=4 \\ \\ y_1=4; \ \ \ y_2 =1 \\ \\ x=5-y \\ \\ x_1=5-4=1; \ \ \ x_2=5-1=4$

$\left \{ {{y_1=4; \ \ \ y_2 = 1 } \atop {x_1=1; \ \ \ x_2 =4 }} \right.$

4.

$\left \{ {{x+y=9} \atop {2^x-2^y=16}} \right. \left \{ {{x=9-y} \atop {2^{9-y}-2^y=16}} \right. \left \{ {{x=9-y} \atop {\frac{2^{9}}{2^y}-2^y=16}} \right. \\ \\ \\ \frac{2^{9}}{2^y}-2^y=16; \ \ \ \frac{512}{2^y}-2^y=16 \ \ \ \cdot | \ 2^y \\ \\ 512-2^{2y}=16\cdot 2^y \\ \\ t=2^y \ (t 0) \\ \\ 512-t^2=16t \\ \\ t^2+16t-512=0 \\ \\ t_{1,2}=\frac{(-\frac{16}{2})\pm \sqrt{(\frac{16}{2})^2- 1 \cdot (-512)}}{1}=-8\pm \sqrt{64+512}=-8 \pm \sqrt{576}=-8\pm 24 \\\\t_1=-8+24=16; \ \ t_2 -8-24=-32$

$2^y=16\\ \\ 2^y=2^4 \\ \\ y=4\\ \\ x=9-y \\ \\ x=9-4=5 \\ \\ \left \{ {{x=5} \atop {y=4}} \right.$