Đáp án+Giải thích các bước giải:

 `c)2x+3\sqrt{x+2}=5`

`(đk : x\ge-2)`

`2x+4+3\sqrt{x+2}-9=5+4-9`

`2.(\sqrt{x+2})^{2}+3\sqrt{x+2}-9=0`

`2.(\sqrt{x+2})^{2}+6\sqrt{x+2}-3\sqrt{x+2}-9=0`

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

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

Do `x\ge-2->\sqrt{x+2}+3\ge3>0AAx`

Nên: `2\sqrt{x+2}-3=0`

`2\sqrt{x+2}=3`

`\sqrt{x+2}=3/2`

`x+2=(3/2)^{2}`

`x+2=9/4`

`x=1/4(tmđk)`

Vậy `S={1/4}`

`d)\sqrt{x+3}+\sqrt{9x+27}+(\sqrt{6x+18})/(\sqrt{6})=5\sqrt{5}`

`(đk : x\ge-3)`

`\sqrt{x+3}+\sqrt{3^{2}.(x+3)}+(\sqrt{6}.\sqrt{x+3})/(\sqrt{6})=5\sqrt{5}`

`\sqrt{x+3}+3\sqrt{x+3}+\sqrt{x+3}=5\sqrt{5}`

`5\sqrt{x+3}=5\sqrt{5}`

`\sqrt{x+3}=\sqrt{5}`

`(\sqrt{x+3})^{2}=(\sqrt{5})^{2}`

`x+3=5`

`x=2(tmđk)`

Vậy `S={2}`

`e)^{**}` `\sqrt{2x+1}+\sqrt{3-2x}=2` `(đk : -1/2\le x\le 3/2)`

Bình phương hai vế ta được:

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

`(\sqrt{2x+1})^{2}+2.\sqrt{(2x+1).(3-2x)}+(\sqrt{3-2x})^{2}=4`

`2x+1+3-2x+2\sqrt{(2x+1).(3-2x)}=4`

`2\sqrt{(2x+1).(3-2x)}=4-4`

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

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

`(2x+1).(3-2x)=0`

`2x+1=0` hoặc `3-2x=0`

`2x=-1` hoặc `2x=3`

`x=-1/2(tmđk)` hoặc `x=3/2(tmđk)`

Vậy `S={-1/2;3/2}`