`a)` `x+3\sqrt{x-1}-5=0`
`<=>3\sqrt{x-1}=5-x` Điều kiện: `1\leqx\leq5`
`<=>(3\sqrt{x-1})^2=(5-x)^2`
`<=>9(x-1)=x^2-10x+25`
`<=>9x-9=x^2-10x+25`
`<=>x^2-10x+25-9x+9=0`
`<=>x^2-19x+34=0`
`<=>x^2-2x-17x+34=0`
`<=>x(x-2)-17(x-2)=0`
`<=>(x-2)(x-17)=0`
`<=>` \(\left[ \begin{array}{l}x-2=0\\x-17=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=2\quad(\text{nhận})\\x=17\quad(\text{loại})\end{array} \right.\)
Vậy `x=2`
