a)
f(x)=\sqrt{4x^2-5x+1}
4x^2-5x+1>0
a=4, b=-5, c=1
\Delta=b^2-4ac=25-4*4*1=25-16=9
\sqrt\Delta=3
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{5-3}{2*4}=\frac{2}{8}=\frac{1}{4}
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{5+3}{8}=1
D_f=(-\infty:\frac{1}{4}\cup(1;+\infty)
http://www.wolframalpha.com/input/?i=4x%5E2-5x%2B1%3E0
b)
f(x)=\sqrt{3-x^2}+\sqrt{x^2-3}
3-x^2\geq0 i x^2-3\geq0
-x^2\geq-3/*(-1) i x^2\geq3
x^2\leq3 i x^2\geq3
x^2=3
x_1=-\sqrt3 , x_2=\sqrt3
D_f= (-\sqrt3;\sqrt3)
c)
f(x)=\frac{2}{\sqrt{-x^2+3x+2}}
-x^2+3x+2>0
a=-1, b=3, c=2
\Delta=9-4*(-1)*2=9+8=17
\sqrt\Delta=\sqrt{17}
x_1=\frac{-3-\sqrt{17}}{-2}=\frac{-(3+\sqrt{17})}{-2}=\frac{3+\sqrt{17}}{2}
x_2=\frac{-3+\sqrt{17}}{-2}=\frac{-(3-\sqrt{17})}{-2}=\frac{3-\sqrt{17}}{2}
D_f= (\frac{3-\sqrt{17}}{2};\frac{3+\sqrt{17}}{2})