`b)` `2\sqrt{9x-27}-2\sqrt{x-3}=1/4\sqrt{16x-48}+9`
Điều kiện: `x\geq3`
`<=>2\sqrt{9(x-3)}-2\sqrt{x-3}=1/4\sqrt{16(x-3)}+9`
`<=>2.\sqrt{9}.\sqrt{x-3}-2\sqrt{x-3}=1/4. \sqrt{16}.\sqrt{x-3}+9`
`<=>6\sqrt{x-3}-2\sqrt{x-3}=\sqrt{x-3}+9`
`<=>4\sqrt{x-3}-\sqrt{x-3}=9`
`<=>3\sqrt{x-3}=9`
`<=>\sqrt{x-3}=3`
`<=>\sqrt{x-3}^2=3^2`
`<=>x-3=9`
`<=>x=9+3`
`<=>x=12` (thoả mãn điều kiện)
Vậy `x=12`
