`19)` `\sqrt{(x-3)^2}=3`
`<=>|x-3|=3`
`<=>` \(\left[ \begin{array}{l}x-3=3\\x-3=-3\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=6\\x=0\end{array} \right.\)
Vậy `S={6;0}`
`20)` `\sqrt{frac{-6}{1+x}}=5` ĐK: `x<` `-1`
`<=>(\sqrt{frac{-6}{1+x}})^2=5^2`
`<=>(-6)/(1+x)=25`
`<=>25(1+x)=-6`
`<=>25+25x+6=0`
`<=>25x=-31`
`<=>x=-(31)/25` (thoả mãn)
Vậy `S={-31/25}`
`22)` `\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1` ĐK: `x\geq-2`
`<=>2\sqrt{x+2}+2\sqrt{x+2}-3\sqrt{x+2}=1`
`<=>\sqrt{x+2}=1`
`<=>(\sqrt{x+2})^2=1^2`
`<=>x+2=1`
`<=>x=-1` (thoả mãn)
Vậy `S={-1}`
`24)` `\sqrt{3x^2-4x+3}=1-2x` ĐK: `x<1/2`
`<=>(\sqrt{3x^2-4x+3})^2=(1-2x)^2`
`<=>3x^2-4x+3=4x^2-4x+1`
`<=>3x^2-4x-4x^2+4x=1-3`
`<=>-x^2=-2`
`<=>x^2=2`
`<=>x=±\sqrt{2}`
Do `x<1/2` nên `x=\sqrt{2}` (loại)
Vậy `S={-\sqrt{2}}`
`25)` `\sqrt{16(x+1)}-\sqrt{9(x+1)}=4` ĐK: `x\geq-1`
`<=>4\sqrt{x+1}-3\sqrt{x+1}=4`
`<=>\sqrt{x+1}=4`
`<=>(\sqrt{x+1})^2=4^2`
`<=>x+1=16`
`<=>x=15` (thoả mãn)
Vậy `S={15}`
