Question

Помогите пожалуйста!!!!!!!!!!!!!!

Answer 1

$\left \{ {{x+y=4} \atop {5xy-x^{2}=-64}} \right.\\\\\\\left \{ {{y=4-x} \atop {5x*(4-x)-x^{2}+64=0 }} \right.\\\\\\\left \{ {{y=4-x} \atop {20x-5x^{2}-x^{2}+64=0 }} \right.\\\\\\\left \{ {{y=4-x} \atop {6x^{2}-20x-64=0 }} \right\\\\\\\left \{ {{y=4-x} \atop {3x^{2}-10x-32=0 }} \right.\\\\3x^{2}-10x-32=0\\\\D=(-10)^{2}-4*3*(-32)=100+384=484=22^{2}\\\\x_{1}=\frac{10-22}{6} =-\frac{12}{6}=-2\\\\x_{2} =\frac{10+22}{6}=\frac{32}{6}=5\frac{1}{3}\\\\y_{1}=4-(-2)=4+2=6 .$

$y_{2}=4-5\frac{1}{3}=-1\frac{1}{3}\\\\Otvet.\boxed{(-2;6),(5\frac{1}{3};-1\frac{1}{3})}$

Answer 2

Ответ:

(7; -3); (-1/3; 4 1/3)

Объяснение:

x+y=4. y=4-x

5xy-x^2=-64. 5x(4-x)-x^2=-64

20x-5x^2-x^2=-64

6x^2-20x-64=0

3x^2-10x-32=0

D=25+96=121

√D=11

x1=[10+11]/3=21/3=7

x2=[10-11]/3= -1/3

y1=4-x1=4-7=-3

y2=4-(-1/3)=4+1/3=4 1/3

(7; -3); (-1/3; 4 1/3)