`\qquad \sqrt{4x+1}+\sqrt{2x+2}=2x-1`
ĐK: `x>=-1/4`
Đặt `\sqrt{4x+1}=a;\sqrt{2x+2}=b` (`a;b>=0`)
Ta thấy `4x+1-(2x+2)=4x+1-2x-2=2x-1`
`=> 2x-1=a^2-b^2`
Ta có pt ẩn phụ `a;b` là:
`\qquad a+b=a^2-b^2`
`<=> a+b=(a-b)(a+b)`
`<=> (a-b)(a+b)-(a+b)=0`
`<=> (a+b)(a-b-1)=0`
`<=> [(a+b=0),(a-b=1):}`
TH1: `a+b=0`
Do `a>=0;b>=0`
`<=> {(a=0),(b=0):}<=>{(4x+1=0),(2x+2=0):}<=>{(x=-1/4),(x=-1):} (\text{vô lý})`
TH2: `a-b=1`
`<=> \sqrt{4x+1}-\sqrt{2x+2}=1`
`<=> \sqrt{4x+1}=1+\sqrt{2x+2}`
`=> 4x+1=1+2x+2+2\sqrt{2x+2}`
`<=> 2\sqrt{2x+2}=2x-2`
`<=>\sqrt{2x+2}=x-1`
`<=> 2x+2=x^2-2x+1\qquad(x>=1)`
`<=> x^2-4x-1=0`
`<=> (x-2)^2=5`
`<=> [(x=2+\sqrt{5}(\text{tm})),(x=2-\sqrt{5}(\text{ktm})):}`
Vậy `S={2+\sqrt{5}}`
