y =
√(4x-x²+12) √(3-x){4x-x²+12≥0{3-x>04x-x²+12≥0x² -4x -12≤0x² -4x -12=0D=16+48=64x₁=
4-8 = -2 2x₂ =
4+8 =6 2 + - +-------------- -2 -------------- 6 -------------------- \\\\\\\\\\\\x∈[-2; 6]3-x>0-x>-3x<3 \\\\\\\\\\\\\\\\\\\\\\\------------- -2 -------- 3 ------------------ 6 ------------- \\\\\\\\\\\\\\\\\\\\\\\\\\\\\x∈[-2; 3)Ответ: 2)