`2(sqrt{9-2x}+sqrt{x-3}-12)=x^2-9x`
`ĐK:3<=x<=9/2`
`pt<=>2sqrt{9-2x}+2sqrt{x-3}-24=x^2-9x`
`<=>x^2-9x-2sqrt{9-2x}-2sqrt{x-3}+24=0`
`<=>9-2x-2sqrt{9-2x}+1+x-3-2sqrt{x-3}+1+x^2-8x+16=0`
`<=>(sqrt{9-2x}-1)^2+(sqrt{x-3}-1)^2+(x-4)^2=0`
Dễ thấy:$\begin{cases}(\sqrt{9-2x}-1)^2 \ge 0\\(\sqrt{x-3}-1)^2 \ge 0\\(x-4)^2 \ge 0\\\end{cases}$
`=>(sqrt{9-2x}-1)^2+(sqrt{x-3}-1)^2+(x-4)^2>=0`
`\text{Mà đề bài cho}:(sqrt{9-2x}-1)^2+(sqrt{x-3}-1)^2+(x-4)^2=0`
`<=>` $\begin{cases}(\sqrt{9-2x}-1)^2=0\\(\sqrt{x-3}-1)^2=0\\(x-4)^2=0\\\end{cases}$
`<=>` $\begin{cases}\sqrt{9-2x}-1=0\\\sqrt{x-3}-1=0\\x-4=0\\\end{cases}$
`<=>` $\begin{cases}9-2x=1\\x-3=1\\x=4\\\end{cases}$
`<=>` $\begin{cases}2x=8\\x=4\\x=4\\\end{cases}$
`<=>x=4(tmđk)`
Vậy phương trình có nghiệm duy nhất `x=4`.
