`b)``\sqrt{4x^2-20x+25}=1`
`<=>\sqrt{(2x-5)^2}=1`
`<=>|2x-5|=1`
`<=> `\(\left[ \begin{array}{l}2x-5=1\\2x-5=-1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=6\\2x=4\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\)
Vậy `S={3;2}`
`d)` ĐKXĐ: $\begin{cases}\sqrt{25x-50}\ge 0\\ \sqrt{x-2}\ge 0\\ \sqrt{9x-18}\ge 0\\\end{cases}$
`=> x\ge 2`
`\sqrt{25x-50}-\frac{\sqrt{x-2}+1}{2}=8\frac{9x-18}{16}`
`<=> \sqrt{25(x-2)}-\frac{\sqrt{x-2}+1}{2}-8.\frac{\sqrt{9(x-2)}}{4}=0`
`=>5.4\sqrt{x-2}-2(\sqrt{x-2}+1)-8.3\sqrt{x-2}=0`
`<=> 20\sqrt{x-2}-2\sqrt{x-2}-2-24\sqrt{x-2}=0`
`<=> -2-6\sqrt{x-2}=0`
`<=>1+3\sqrt{x-2}=0`
`<=> 3\sqrt{x-1}=-1 ` (loại)
`=>` Phương trình vô nghiệm
Vậy `S={∅}`
