х(3х-1)-х^2+16 ≤ х(2-х) -х(11-2х)3х² - х - х² + 16 ≤ 2х -х² - 11х+ 2х²2х² - х +16 -2х + х² +11х - 2х² ≤ 0 х² + 8х +16 ≤ 0
x2 + 8x + 16 = 0D = b2 - 4acD = 64 - 64 = 0
x = - b/2ax = - 8/2= -4ответ: x = -4
х∈ [-4, 0]
х(3х-1)-х^2+16 ≤ х(2-х) -х(11-2х)
3х² - х - х² + 16 ≤ 2х -х² - 11х+ 2х²
2х² - х +16 -2х + х² +11х - 2х² ≤ 0
х² + 8х +16 ≤ 0
x2 + 8x + 16 = 0
D = b2 - 4ac
D = 64 - 64 = 0
x = - b/2a
x = - 8/2= -4
ответ: x = -4
х∈ [-4, 0]