Bonjour,
1)
[tex] f(x)=2x^3-x^2-1\\dom f=\mathbb{R}\\f(x)=x(2x^2-x-1)=x(2x^2+x-2x-1)\\=x(x(2x+1)-1(2x+1))\\=x(2x+1)(x-1)\\Zero=\{-\frac{1}{2} ,0,1\}\\ [/tex]
2)
[tex] g(x)=\sqrt{3x-2} \\3x-2\geq 0\Rightarrow\ x\geq \frac{2}{3} \\Dom\ g=[\frac{2}{3}\ \infty)\\3x-2=0\ \Rightarrow\ x=\frac{2}{3} \\Zero=\{ \frac{2}{3}\}\\ [/tex]
3)
[tex] h(x)=\dfrac{(x+1)(2x-3)(2-x)}{x-3} \\Dom\ h=\mathbb{R}-\{3\}\\Zero={-1,\frac{3}{2},2\} \\ [/tex]
4)
[tex] i(x)=\sqrt{\dfrac{x^2-3x+2}{x-4}} = \sqrt{\dfrac{(x-1)(x-2)}{x-4}} \\1)\ x\neq 4\\2)\\\begin{array}{l|ccccccc}&&1&&2&&4&\\x-1&-&0&+&+&+&+&+\\x-2&-&-&-&0&+&+&+\\x-4&-&-&-&-&-&0&+\\\\\dfrac{(x-1)(x-2)}{x-4}&-&0&+&0&-&\zeta&+\\\end{array}\right]\\\\dom\ i=[1\ 2]\ U\ ]4\ \infty)\\Zero=\{1,2\} \\ [/tex]
