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

ОДЗ {2x²-x>0⇒x(2x-1)>0⇒x<0 U x>1/2 {x+4>0⇒x>-4 {x+4≠1⇒x≠-3 x∈(-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)≤0 log(2)(2x²-x)*(log(2)(x+4)-1)≤0 1){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≤-2 x≤-2 x∈(-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≤1 x∈[-1/2;0) U (1/2;1] Ответ x∈(-4;-3) U (-3;-2] U [-1/2;0) U (1/2;1]