Так?
(320)^(1/2)=16^(1/2)*4^(1/2)*5^(1/2)=4*2*5(1/2)=8*5^(1/2) (восемь * корень из 5)
Xу - 2х + 3у = 6
2ху - 3х + 5у = 11
xy - 2x = 6 - 3y
x (y - 2) = 6 - 3y
x = (6-3y)/(y-2)
2y (6-3y)/(y-2) - 3 (6-3y)/(y-2) + 5y = 11
(12y - 6y2) / (y - 2) - ( 18 - 9y )/ (y-2) + 5y = 11
12y - 6y2 - 18 + 9y + 5y (y-2) = 11 (y-2)
12y - 6y2 - 18 + 9y + 5y2 - 10y = 11y - 22
12y + 9y - 10y - 11y - 6y2 + 5y2 - 18 + 22 = 0
12y + 9y - 10y - 11y - 6y2 + 5y2 - 18 + 22 = 0
0y - y2 + 4 = 0
y2 = 4
ищем Х:
x = (6-3y)/(y-2)
x1 = (6 - 3 * 2) / (2 - 2) - на ноль делить нельзя
x2 = (6 - 3 * -2) / (-2 - 2) = 6 +6 / -4 = 12 / -4 = -3
ответ только 1:
y = -2
х = -3
=2(5m-5)/-2=-(5m-5)=-5m+5
task/29460089
Пусть f(x ) +f(y) = f(z ) , f(x) = Log (1+x) / (1 - x)
z → ?
Log (1+x) / (1 - x) + Log (1+y) / (1 - y) = Log (1+z) / (1 - z) ;
* * * x ; y ; z ∈ (-1 ; 1) * * *
Log (1+x)(1+y) /(1 -x)(1 -y) = Log (1+z) / (1 - z) ;
(1+x)(1+y) / (1 -x)(1 -y) = (1+z) / (1 - z );
(1+x)(1+y) (1 -z) ) = (1 - x )(1 -y)( 1+z) ;
( (1 - x )(1 -y) + (1+x)(1+y) )* z =(1+x)(1+y) - (1- x)(1- y ) ;
2(1+xy) *z = 2(x+y) ;
z = (x+y) / (xy +1) . xy ≠ - 1