Question

cine ma ajuta la primele doua exercitii ???
repedeeeee

Answer 1

1)

[tex]\it \sqrt{\dfrac{361}{16}} =\sqrt{\dfrac{19^2}{4^2}}=\dfrac{19}{4} \in \mathbb{Q} \\\;\\ \\\;\\ \sqrt{6\dfrac{97}{144}} =\sqrt{\dfrac{961}{144}} =\sqrt{\dfrac{19^2}{12^2}} =\dfrac{19}{12} \in \mathbb{Q} [/tex]

[tex]\it A\cap\mathbb{Q} =\left \{-2,5;\ -3,5(6);\ \sqrt{\dfrac{361}{16}};\ \sqrt{6\dfrac{97}{144}}\right\} \\\;\\ \\\;\\ A\cap(\mathbb{R} \backslash\mathbb{Q}}) =\left\{\sqrt{2\dfrac{3}{4}};\ \sqrt{17}; \ -2\sqrt3;\ \sqrt5\right\}[/tex]

2)

[tex]\it \sqrt{3^2+4^2} =\sqrt{9+16} =\sqrt{25} =5 \in \mathbb{Q} \\\;\\ \\\;\\ \sqrt{11+\dfrac{1}{9}} =\sqrt{\dfrac{100}{9}} =\dfrac{10}{3} \in \mathbb{Q} \\\;\\ \\\;\\ -\dfrac{\sqrt{289}}{3} =-\dfrac{17}{3} \in \mathbb{Q}[/tex]

$\it A\cap\mathbb{Q} =\left\{-7,3;\ \sqrt{3^2+4^2};\ -5,(2); \sqrt{11+\dfrac{1}{9}};\ -\dfrac{\sqrt{289}}3{}\right\}$