`***`Lời giải`***`
a)
`\sqrt{36x^2-12x+1}=5`
`<=>1-12x+36x^2=25`
`<=>36x^2-12x-24=0`
`<=>12(3x^2-x-2)=0`
`<=>3x^2-x-2=0`
`<=>3x^2-3x+2x-2=0`
`<=>3x(x-1)+2(x-1)=0`
`<=>(3x+2)(x-1)=0`
`<=>`\(\left[ \begin{array}{l}x-1=0\\3x+2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1\\3x=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1\\x=-2/3\end{array} \right.\)
Vậy `S={1;-2/3}`
b)
ĐK: `x≤4/5`
`\sqrt{4-5x}=12`
`<=>4-5x=144`
`<=>-5x=144-4`
`<=>-5x=140`
`<=>x=-28(N)`
Vậy `S={-28}`
c)
`\sqrt{9x^2}=2x+1`
`<=>\sqrt{(3x)^2}=2x+1`
`<=>|3x|=2x+1`
`<=>`$\begin{cases} 3x=2x+1(x≥0)\\-3x=2x+1(x<0) \end{cases}$
`<=>`$\begin{cases} 3x-2x=1\\-3x-2x=1 \end{cases}$
`<=>`$\begin{cases} x=1(N)\\x=-1/5(N) \end{cases}$
Vậy `S={1;-1/5}`
