Question

log3(x^2-x-3)+log3(2x^2+x-3)>=log3(x^2-2)^2+2+log1/3 4

Answer 1

ОДЗ
x²-x-3>0
D=1+12=13
x1=(1-√13)/2 U x2=(1+√13)/2
x<(1-√13)/2 U x>(1+√13)/2
2x²+x-3>0
D=1+24=25
x1=(-1-5)/4=-1,5 U x2=(-1+5)/4=1
x<-1,5 U x>1
x²-2≠0⇒x≠-√2 U x≠√2
x∈(-∞;-1,5) U ((1+√13)/2;∞)
log(3)[(x²-x-3)(2x²+x-3)]≥log(3)[9(x²-2)²/4]      2=log(3)9 U log(1/3)4=-log(3)4
(x²-x-3)(2x²+x-3)≥9(x²-2)²/4
4(x²-x-3)(2x²+x-3)≥9(x²-2)²
8x^4+4x³-12x²-8x³-4x²+12x-24x²-12x+36-9x^4+36x²-36≥0
-x^4-4x³-4x²≥0
x^4+4x³+4x²≤0
x²(x²+4x+4)²≤0
x²(x+2)²≤0
x=0 ∉ ОДЗ
 x=-2
Ответ х=-2
Проверка
log(3)(4+2-3)+log(3)(8-2-3)=log(3)3+log(3)3=1+1=2
log(3)(4-2)²+2-log(3)4=log(3)4+2-log(3)4=2
2≥2