Question

решите систему уравнений
log4 (x+y)=2
log3 x + log3 y = 2 + log3 7

Answer 1

x + y = 4²
x + y = 16
log3 x + log3 y = 2 + log3 7

2 = log3 9
log3 x + log3 y = log3 9 + log3 7
log3 (x*y) = log3 (9*7)
x*y = 63
x=63/y

x + y = 16     63/y + y -16 = 0
(63/y + y -16 = 0)*y
63 + y² - 16y = 0
D = (b)²-4*a*c = (-16)² - 4*1*63 = 256 - 252 = 4
y1,2 = (-b + & - √D)/2a
y1 = (16 - 2)/2*1 = 14/2 = 7
y2= (16 + 2)/2 = 18/2 = 9

x1 + y1 = 16
x1 = 16 -y1 = 16 - 7 = 9
x1 = 9,
x2 + y2 = 16
x2 = 16 - y2 = 16 - 9 = 7
x2 = 7.
Ответ (7 ; 9),(9 ; 7)