у=
√(14+5х-х²) +
3 3 √(20+х-х²){14+5x-x² ≥0{20+x-x² >014+5x-x² ≥0x²-5x-14≤0x²-5x-14=0D=25+14*4=25+56=81x₁=
5-9= -2 2x₂=
5+9=7 2 + - +-------- -2 ----------- 7 --------- \\\\\\\\\\\x∈[-2; 7]20+x-x² >0x²-x-20 <0x²-x-20=0D=1+4*20=81x₁=
1-9=-4 2x₂=
1+9=5 2 + - +-------- -4 --------- 5 ------------- \\\\\\\\\\\\x∈(-4; 5){x∈[-2; 7]{x∈(-4; 5) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\--------- -4 -------- -2 ---------- 5 ---------- 7 ----------- \\\\\\\\\\\\\\\\\\\\\\\\\\\\\x∈[-2; 5)D(y)=[-2; 5) - область определения функции.