`\sqrt{x-1}+\sqrt{2x-1}=5` `(x>=1)`
⇔`\sqrt{x-1}+\sqrt{2x-1}-5=0`
⇔`(\sqrt{x-1}-2)+(\sqrt{2x-1}-3)=0`
⇔`((\sqrt{x-1}-2)(\sqrt{x-1}+2))/(\sqrt{x-1}+2)+((\sqrt{2x-1}-3)(\sqrt{2x-1}+3))/(\sqrt{2x-1}+3)=0`
⇔`(x-1-4)/(\sqrt{x-1}+2)+(2x-1-9)/(\sqrt{2x-1}+3)=0`
⇔`(x-5)/(\sqrt{x-1}+2)+(2x-10)/(\sqrt{2x-1}+3)=0`
⇔`(x-5)/(\sqrt{x-1}+2)+(2(x-5))/(\sqrt{2x-1}+3)=0`
⇔`(x-5)(1/(\sqrt{x-1}+2)+2/(\sqrt{2x-1}+3))=0`
Vì `1/(\sqrt{x-1}+2)+2/(\sqrt{2x-1}+3)ne0`
⇒`x-5=0`
⇔`x=5(tm)`
Vậy `S={5}`
