Resposta:
Letra E.
Explicação passo-a-passo:
[tex]f(x)=\left \bigg\{ \big{{2x,\forall x\in\mathbb{Q}} \atop {1+x^2,\forall x\in\mathbb{I}}} \right.[/tex]
[tex]\left\{0\right\}\in\mathbb{Q}\ \therefore\ f(0)=2(0)\ \therefore\ \boxed{f(0)=0}[/tex]
[tex]\left\{\dfrac{3}{4}\right\}\in\mathbb{Q}\ \therefore\ f\bigg(\dfrac{3}{4}\bigg)=2\bigg(\dfrac{3}{4}\bigg)\ \therefore\ \boxed{f\bigg(\dfrac{3}{4}\bigg)=\dfrac{3}{2}}[/tex]
[tex]\left\{1-2\sqrt{2}\right\}\in\mathbb{I}\ \therefore\ f(1-2\sqrt{2})=1+(1-2\sqrt{2})^2=\\\\ =1+1-4\sqrt{2}+8\ \therefore\ \boxed{f(1-2\sqrt{2})=10-4\sqrt{2}}[/tex]
Dessa forma, o valor da expressão será:
[tex]f(0)+f\bigg(\dfrac{3}{4}\bigg)-f(1-2\sqrt{2})=\\\\ =0+\dfrac{3}{2}-(10-4\sqrt{2})=\dfrac{3}{2}-\dfrac{20}{2}+4\sqrt{2}\ \therefore[/tex]
[tex]f(0)+f\bigg(\dfrac{3}{4}\bigg)-f(1-2\sqrt{2})=\boxed{-\dfrac{17}{2}+4\sqrt{2}}[/tex]
