`c) x-sqrt{x}-12=0`
`<=> x-4sqrt{x}+3sqrt{x}-12=0`
`<=> sqrt{x}(sqrt{x}-4)+3(sqrt{x}-4)=0`
`<=> (sqrt{x}-4)(sqrt{x}+3)=0`
`=> sqrt{x}-4=0` $(\text{do $\sqrt{x}+3>0$ với mọi $x\ge0$})$
`<=> sqrt{x}=4`
`<=> x=16 (\text{tm})`
Vậy `S={16}`
`d) x sqrt{x-1}+2sqrt{x+2}=2x+sqrt{x^2+x-2}` ĐK:`x>=1`
Đặt `sqrt{x-1}=a; sqrt{x+2}=b\quad(a>=0;b>0)`
`-> sqrt{x^2+x-2}=sqrt{x-1}.sqrt{x+2}=ab`
Khi đó ta có pt ẩn phụ `a;b` là:
`\qquad ax+2b=2x+ab`
`<=> ax+2b-2x-ab=0`
`<=> x(a-2)-b(a-2)=0`
`<=> (a-2)(x-b)=0`
`<=> [(a=2),(b=x):} (\text{tm})`
Khi `a=2`
`-> sqrt{x-1}=2`
`<=> x-1=4`
`<=> x=5 (\text{tm})`
Khi `b=x`
`-> sqrt{x+2}=x`
`<=> x+2=x^2`
`<=> x^2-x-2=0`
`<=> x^2+x-2x-2=0`
`<=> (x+1)(x-2)=0`
`<=> [(x=-1(\text{ktm})),(x=2(\text{tm})):}`
Vậy `S={5;2}`
