Đáp án + Giải thích các bước giải:
14a) `x^2-3=x^2-(\sqrt3)^2=(x-\sqrt3)(x+\sqrt3)`
14b) `x^2-6=x^2-(\sqrt6)^2=(x-\sqrt6)(x+\sqrt6)`
14c) `x^2+2\sqrt3x+3=x^2+2.x.\sqrt3+(\sqrt3)^2=(x+\sqrt3)^2`
14d) `x^2-2\sqrt5x+5=x^2-2.x.\sqrt5+(\sqrt5)^2=(x-\sqrt5)^2`
15a) `x^2-5=0`
`=>x^2-(\sqrt5)^2=0`
`=>(x-\sqrt5)(x+sqrt5)=0`
`=>x-\sqrt5=0` hoặc `x+\sqrt5=0`
`=>x=\sqrt5` hoặc `x=-\sqrt5`
Vậy `S={\pm\sqrt5}`
15b) `x^2-2\sqrt{11}x+11=0`
`=>x^2-2.x.\sqrt{11}+(\sqrt11)^2=0`
`=>(x-\sqrt{11})^2=0`
`=>x-\sqrt{11}=0`
`=>x=\sqrt{11}`
Vậy `S={\sqrt{11}}`
