`a)`
`(x-1)^2=2x-3`
`<=>x^2-2x+1=2x-3`
`<=>x^2-2x+1-2x+3=0`
`<=>x^2-4x+4=0`
`<=>x^2-2.x.2+2^2=0`
`<=>(x-2)^2=0`
`<=>x-2=0`
`<=>x=2`
Vậy `S={2}`
`b)` Tìm ĐKXĐ :
`{(x+2ge0),(6-xge0):}`
`<=>{(xge-2),(xle6):}`
`<=>-2lexle6`
`\sqrt{x+2}=\sqrt{6-x}` Điều kiện : `-2lexle6`
`<=>(\sqrt{x+2})^2=(\sqrt{6-x})^2`
`<=>x+2=6-x`
`<=>x+x=6-2`
`<=>2x=4`
`<=>x=2 ` ( thỏa mãn )
Vậy `S={2}`