Question

Помогите пожалуйста алгебра. С.м работа

Answer 1

$1. left { {{x^2-5y^2}=11 atop {x^2+5y^2=61}}+ right. \ ------- \ left { {{} 2x^2=72;|:2 }} right. left { {{} x^2=36; }} right. ; \ x_1=6; x_2=-6 = textgreater \ (-6)^2-5y^2=11; 36-5y^2=11; -5y^2=-25; y_1=5, y_2=-5; \ left { {{x_{1,2}=+-6} atop {y_{1:2}=+-5}} right. \ 2. left { {{x+y=6} atop {xy=8}} right. ; left { {{x=-y+6} atop {(-y+6)*y=8}}; right. ; \ -y^2-6y-8=0; |*(-1); y^2+6y+8=0 ; (y+2)(y+4)=0; \ y_1=-2 ; y_2=-4; \  ]x_1=-(-2)+6; x_1=8; x_2=-(-4)+6; x_2=10; \ left { {{x_{1}=8, x_2=10 atop {y_1=-2, y_2=-4}} right. ;$
$left { {{x+y=-2} atop {x-y=-8}}+ right. \ ----- \ left { {{2x=-10; } right.; \ left x=-5; right. \ -5-y=-8; -y=-3; y=3; \ left { {{x=-5} atop {y=3}} right.  left { {{x^2-y^2=6} atop {x^2+y^2=10}}+ right. \ ----- \ 2x^2=16; \ x^2=8; \ x_{1.2}=-+ sqrt{8}; \ ( sqrt{8})^2-y^2=6; \ -y^2=-2; \ y_{1,2}=+- sqrt{2}; \ left { {{x_{1,2}=+- sqrt{5} } atop {y_{1,2}=+- sqrt{2}} right. \ left { {{3x+y=2} atop {x^2-xy+6y=-4}} right.; left { {{y=-3x+2} atop {(-3x+2)^2-x(-3x+2)}+6(-3x+2)=-4} right. ; \ 12x^2-32+16=-4; \ 12x^2-32+20=0;|:2 \ 6x^2-16+10=0; \ D=b^2-4ac; 16^2-4*6*10; 256-240=16, D textgreater 0; \ x_{1,2}= frac{-b+- sqrt{D} }{2a}; \ x_1= frac{16+4}{12}; x_1= 1frac{2}{3}; \ x_2= frac{12}{12}=1; \ 3*( frac{5}{3})+y=2; \ y_1=-3; \ 3*1+y=2; \ y_2=-1; \ left { {{x_{1}=1 frac{2}{3} };x_2=1 atop {y_1=-3; y_2=-1}} right.$

Answer 2

A - не пишите, это ошибка при вводе кода.