Đáp án:

 `S={-1}`

Giải thích các bước giải:

$\color{#a8b0b4}{⸻}\color{#91999d}{⸻}\color{#7a8286}{⸸}\color{#636b6f}{⛧}
\color{#4c5356}{𝕯}\color{#434a4d}{ψ}\color{#3b4244}{ภ}\color{#32393b}{ค}
\color{#2a3133}{๓}\color{#21282a}{๏}\color{#1a2123}{ภ}\color{#141a1c}{ร}
\color{#0f1516}{ฬ}\color{#0a1011}{๏}\color{#05090a}{ɭ}\color{#020406}{๔}
\color{#05090a}{⛧}\color{#0a1011}{⸸}\color{#141a1c}{⸻}\color{#1a2123}{⸻}$

 `c)\sqrt{1-x^{2}}+\sqrt{x^{2}+3x+2}=x+1`

`(đk: -1\le x\le1)`

`<=>\sqrt{1^{2}-x^{2}}+\sqrt{x^{2}+x+2x+2}=x+1`

`<=>\sqrt{(1-x).(1+x)}+\sqrt{x.(x+1)+2.(x+1)}=(\sqrt{x+1})^{2}`

`<=>\sqrt{1-x}.\sqrt{1+x}+\sqrt{(x+1).(x+2)}-\sqrt{x+1}.\sqrt{x+1}=0`

`<=>\sqrt{x+1}.(\sqrt{1-x}+\sqrt{x+2}-\sqrt{x+1})=0`

`+) TH1: \sqrt{x+1}=0`

`<=>x+1=0^{2}`

`<=>x=-1(tmđk)`

`+)TH2: \sqrt{1-x}+\sqrt{x+2}-\sqrt{x+1}=0`

`<=>\sqrt{1-x}+\sqrt{x+2}=\sqrt{x+1}`

`<=>(\sqrt{1-x}+\sqrt{x+2})^{2}=x+1`

`<=>1-x+x+2+2\sqrt{(1-x).(x+2)}=x+1`

`<=>3+2\sqrt{(1-x).(x+2)}-x-1=0`

`<=>2\sqrt{(1-x).(x+2)}+2-x=0`

`<=>2\sqrt{(1-x).(x+2)}=x-2`

`<=>[2\sqrt{(1-x).(x+2)}]=(x-2)^{2}`

`<=>x^{2}-2.x.2+2^{2}=4.(1-x).(x+2)`

`<=>x^{2}-4x+4=-4.(x-1).(x+2)`

`<=>x^{2}-4x+4=-4.(x^{2}+2x-x-2)`

`<=>x^{2}-4x+4+4.(x^{2}+x-2)=0`

`<=>x^{2}-4x+4+4x^{2}+4x-8=0`

`<=>5x^{2}-4=0`

`<=>x^{2}=4/5`

`<=>x^{2}=(\pm\frac{2}{\sqrt{5}})^{2}`

`<=>x=-\frac{2}{\sqrt{5}}(ktmđk)` hoặc `x=\frac{2}{\sqrt{5}}(ktmđk)`

`->` `TH2` ta loại

Vậy `S={-1}`