Đáp án:
`S={7/9}.`
Giải thích các bước giải:
`2x-3+2\sqrt{x^2+3x+2}=\sqrt{x+1}+\sqrt{x+2}(x>=-1)`
Đặt `\sqrt{x+1}+\sqrt{x+2}=a(a>0)`
`<=>a^2=x+1+x+2+2\sqrt{(x+1)(x+2)}`
`<=>a^2=2x+3+2\sqrt{(x+1)(x+2)}`
`pt<=>2x+3+2\sqrt{(x+1)(x+2)}-6=\sqrt{x+1}+\sqrt{x+2}`
`<=>a^2-6=a`
`<=>a^2-a-6=0`
`<=>(a+2)(a-3)=0`
`a>0=>a+2>2>0`
`<=>a-3=0<=>a=3`
`<=>\sqrt{x+1}+\sqrt{x+2}=3`
`<=>(\sqrt{x+1}+\sqrt{x+2})^2=9`
`<=>x+1+x+2+2\sqrt{(x+1)(x+2)}=9`
`<=>2\sqrt{x^2+3x+2}=6-2x`
`<=>\sqrt{x^2+3x+2}=3-x`
Vì `VT>=0=>3-x>=0<=>x<=3=>-1<=x<=3(đkxđ)`
`<=>x^2+3x+2=(3-x)^2`
`<=>x^2+3x+2=x^2-6x+9`
`<=>9x=7`
`<=>x=7/9(tmđk)`
Vậy `S={7/9}.`
