`\sqrt{16x^2+24x+9}-5=0`

`\to \sqrt{(4x+3)^2}=5`

`\to |4x+3|=5`

`\to` \(\left[ \begin{array}{l}4x+3=5\\4x+3=-5\end{array} \right.\) `\to` \(\left[ \begin{array}{l}4x=2\\4x=-8\end{array} \right.\) `\to` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-2\end{array} \right.\) 

Vậy `x\in{-2;1/2}`

`\sqrt{x^2-4x+4}=7`

`\to \sqrt{(x-2)^2}=7`

`\to |x-2|=7`

`\to` \(\left[ \begin{array}{l}x-2=7\\x-2=-7\end{array} \right.\) `\to` \(\left[ \begin{array}{l}x=9\\x=-5\end{array} \right.\) 

Vậy `x\in{-5;9}`

`\sqrt{x^2-6x+9}=x+5(x\ge-5)`

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

`\to |x-3|=x+5`

`\to` \(\left[ \begin{array}{l}x-3=x+5\\x-3=-x-5\end{array} \right.\) `\to` \(\left[ \begin{array}{l}0x=8(L)\\2x=-2\to x=-1(TM)\end{array} \right.\) 

Vậy `x=-1`

`a)` `\sqrt{16x^2+24x+9}-5=0`

`<=>\sqrt{(4x+3)^2}=5`

`<=>|4x+3|=5`

`<=>` \(\left[ \begin{array}{l}4x+3=5\\4x+3=-5\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}4x=2\\4x=-8\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-2\end{array} \right.\) 

Vậy `S={1/2;-2}`

`b)` `\sqrt{x^2-4x+4}=7`

`<=>\sqrt{(x-2)^2}=7`

`<=>|x-2|=7`

`<=>` \(\left[ \begin{array}{l}x-2=7\\x-2=-7\end{array} \right.\) \(\left[ \begin{array}{l}x=9\\x=-5\end{array} \right.\) 

Vậy `S={9;-5}`

`c)` `\sqrt{x^2-6x+9}=x+5`   ĐKXĐ: `x\geq-5`

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

`<=>|x-3|=x+5`

`<=>` \(\left[ \begin{array}{l}x-3=x+5\\x-3=-(x+5)\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x-x=5+3\\x-3=-x-5\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}0x=8\\x+x=-5+3\end{array} \right.\)`<=>`\(\left[ \begin{array}{l}x∉R\\2x=-2\end{array} \right.\)`<=>x=-1` `(TMĐK)`

Vậy `x=-1`