`a)sqrt{x^2+4}=x-2``đk:x-2>=0 x>=2`BP 2 vế ta có:`x^2+4=x^2-4x+4`` 4x=0`` x=0(l)`Vậy pt vô nghiệm`b)sqrt{x^2-10x+25}=3-19x``đk:3-19x>=0 x \sqrt{(x-5)^2}=3-19x`` |x-5|=3-19x``+)x-5=3-19x`` 20x=8 x=2/5(tm)``+)x-5=19x-3`` 18x=-2`` x=-1/9(tm)`Vậy `S={2/5,-1/9}``c)sqrt{x^2-9}+sqrt{x^2-6x+9}=0``đk:x^2-9>=0 x^2>=9`Vì `sqrt{x^2-9}>=0``sqrt{x^2-6x+9}>=0``=>sqrt{x^2-9}=sqrt{x^2-6x+9}=0`` sqrt{(x-3)(x+3)}=sqrt{(x-3)^2}=0`` x=3`Vậy `S={3]`Câu trả lời: `a)sqrt{x^2+4}=x-2``đk:x-2>=0 x>=2`BP 2 vế ta có:`x^2+4=x^2-4x+4`` 4x=0`` x=0(l)`Vậy pt vô nghiệm`b)sqrt{x^2-10x+25}=3-19x``đk:3-19x>=0 x \sqrt{(x-5)^2}=3-19x`` |x-5|=3-19x``+)x-5=3-19x`` 20x=8 x=2/5(tm)``+)x-5=19x-3`` 18x=-2`` x=-1/9(tm)`Vậy `S={2/5,-1/9}``c)sqrt{x^2-9}+sqrt{x^2-6x+9}=0``đk:x^2-9>=0 x^2>=9`Vì `sqrt{x^2-9}>=0``sqrt{x^2-6x+9}>=0``=>sqrt{x^2-9}=sqrt{x^2-6x+9}=0`` sqrt{(x-3)(x+3)}=sqrt{(x-3)^2}=0`` x=3`Vậy `S={3]`