Log√2x 4 log4(3x-2) 3 log√6(x-4)+log√6(x+1)≤2 - вопрос №244323646126025119 от innarudenkoco 07.08.2020 00:34

Question

Алгебра: Log√2x 4 log4(3x-2) 3 log√6(x-4)+log√6(x+1)≤2

innarudenkoco · Answer

1log(√2)x 4x (√2)^4x 4x∈(4;∞)2{3x-2 0⇒3x 2⇒x 2/3{3x-2 64⇒3x 66⇒x 22x∈(2/3;22)3{x-4 0⇒x 4{x+1 0⇒x -1{log(√6)[(x-4)(x+1)]≤2⇒x²-3x-4≤6x²-3x-10≤0x1+x2=3 U x1*x2=-10⇒x1=-2 U x2=5-2≤x≤5 U x 4⇒x∈(4;5]...

Log√2x> 4 log4(3x-2)< 3 log√6(x-4)+log√6(x+1)≤2