(x-3)*√(x2+4)<=x2-9;
(x-1)*√(x2+1)<=x2-1;
помогите решить плиз)

1ОДЗx²+4≥0⇒x∈R(x-3)√(x²+4)-(x-3)(x+3)≤0(x-3)*(√(x²+4)-(x+3))≤01){x-3≤0⇒x≤3{√(x²+4)-(x+3)≥0⇒√(x²+4)≥x+3x²+4≥x²+6x+96x≤-5x≤-5/6x∈(-∞;-5/6]2)x≥3x≥-5/6x∈[3;∞)Ответ x∈(-∞;-5/6] U [3;∞)2ОДЗ x²+1≥0⇒x∈R(x-1)√(x²+1)-(x-1)(x+1)≤0(x-1)*(√(x²+4)-(x+1))≤01){x-1≤0⇒x≤1{√(x²+1)-(x+1)≥0⇒√(x²+1)≥x+1x²+1≥x²+2x+12x≤0x≤0x∈(-∞;0]2)x≥1x≥0x∈[1;∞)Ответ x∈(-∞;0] U [1;∞)