Question

 срочно,нужно решить систему
(2x+4)^2=3y
(4x+2)^2=3y

Answer 1

$left { {{(2x+4)^2=3y} atop {(4x+2)^2=3y}} right.\ \ (2x+4)^2=(4x+2)^2\ 4x^2+16x+16=16x^2+16x+4\ 4x^2+16x+16-16x^2-16x-4=0\ -12x^2 +12=0\ -12x^2=-12 |:(-12)\ x^2=1\ \ left[begin{array}{ccc}x_1=1\x_2=-1end{array}right \ y_1=frac{(2x_1+4)^2}3=frac{(2*1+4)^2}{3}=frac{36}3=12\ y_2=frac{(2x_2+4)^2}3=frac{(2*(-1)+4)^2}{3}=frac{4}3$
Ответ: (1; 12) или (-1; 4/3)

Answer 2

и названия?

Answer 3

обнови страницу

Answer 4

есть?

Answer 5

да,спасибо)

Answer 6

ок!!!

Answer 7

$left { {{(2x+4)^2=3y,} atop {(4x+2)^2=3y;}} right. left { {{(2x+4)^2=3y,} atop {(4x+2)^2=(2x+4)^2;}} right. left { {{(2x+4)^2=3y,} atop { left [ {{4x+2=2x+4,} atop {4x+2=-(2x+4);}} right.}} right. left [ {{ left { {{(2x+4)^2=3y,} atop {2x=2;}} right. } atop { left { {{(2x+4)^2=3y,} atop {6x=-6;}} right.}} right.$
$left [ {{ left { {{x=1,} atop {3y=36;}} right. } atop { left { {{x=-1,} atop {3y=4;}} right.}} right. left [ {{ left { {{x=1,} atop {y=12;}} right. } atop { left { {{x=-1,} atop {y=1 frac{1}{3} ;}} right.}} right. \ (1;12), (-1;1 frac{1}{3}).$