Đáp án: `S={19;20}.`
Giải thích các bước giải:
`\sqrt{x-19}+\sqrt{20-x}=1+\sqrt{(x-19)(20-x)}`
Điều kiện:`{(x-19>=0),(20-x>=0):}`
`<=>{(x>=19),(x<=20):}`
`<=>19<=x<=20`
`pt<=>\sqrt{(x-19)(20-x)}-\sqrt{x-19}+1-\sqrt{20-x}=0`
`<=>\sqrt{x-19}(\sqrt{20-x}-1)-(\sqrt{20-x}-1)=0`
`<=>(\sqrt{20-x}-1)(\sqrt{x-19}-1)=0`
`<=>[(\sqrt{20-x}-1=0),(\sqrt{x-19}-1=0):}`
`<=>[(\sqrt{20-x}=1),(\sqrt{x-19}=1):}`
`<=>[(20-x=1),(x-19=1):}`
`<=>[(x=19(TM)),(x=20(TM)):}`
Vậy `S={19;20}.`
