Log3(x^2-x-3)+log3(2x^2+x-3) =log3(x^2-2)^2+2+log1/3 - вопрос №3974048323322332222134 от ervinbek2005 02.01.2023 11:54

Question

Математика: Log3(x^2-x-3)+log3(2x^2+x-3) =log3(x^2-2)^2+2+log1/3 4

ervinbek2005 · Answer

ОДЗx²-x-3 0D=1+12=13x1=(1-√13)/2 U x2=(1+√13)/2x (1-√13)/2 U x (1+√13)/22x²+x-3 0D=1+24=25x1=(-1-5)/4=-1,5 U x2=(-1+5)/4=1x -1,5 U x 1x²-2≠0⇒x≠-√2 U x≠√2x∈(-∞;-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)²/44(x²-x-3)(2x²+x-3)≥9(x²-2)²8x^4+4x³-12x²-8x³-4x²+12x-24x²-12x+36-9x^4+36x²-36≥0-x^4-4x³-4x²≥0x^4+4x³+4x²≤0x²(x²+4x+4)²≤0x²(x+2)²≤0x=0 ∉ ОДЗ x=-2ответ х=-2Проверкаlog(3)(4+2-3)+log(3)(8-2-3)=log(3)3+log(3)3=1+1=2log(3)(4-2)²+2-log(3)4=log(3)4+2-log(3)4=22≥2...

Log3(x^2-x-3)+log3(2x^2+x-3)> =log3(x^2-2)^2+2+log1/3 4