Решите неравенство
[tex](x+4) ^{log_{2}(2x^2-x) } \leq 2x^2-x[/tex]

ОДЗ{2x²-x>0⇒x(2x-1)>0⇒x<0 U x>1/2{x+4>0⇒x>-4{x+4≠1⇒x≠-3x∈(-4;-3) U (-3;0) U (1/2;∞)-------------------------log(2)[(x+4)^log(2)(2x²-x))≤log(2)(2x²-x)log(2)(x+4)*log(2)(2x²-x)-log(2)(2x²-x)≤0log(2)(2x²-x)*(log(2)(x+4)-1)≤01){log(2)(2x²-x)≥0{log(2)(x+4)-1≤0{(2x²-x)≥1⇒2x²-x-1≥0  D=1+8=9 x1=-1/2 U x2=1⇒x≤-1/2 U x≥1{log(2)(x+4)≤1⇒x+4≤2⇒x≤-2x≤-2x∈(-4;-3) U (-3;-2]2){log(2)(2x²-x)≤0{log(2)(x+4)-1≥0{-1/2≤x≤1{x≥-2-1/2≤x≤1x∈[-1/2;0) U (1/2;1]Ответ x∈(-4;-3) U (-3;-2] U [-1/2;0) U (1/2;1]