\(a,f\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{2}\right)^2+4=\dfrac{1}{4}+4=\dfrac{17}{4}\\ f\left(5\right)=5^2+4=25+4=29\\ b,f\left(x\right)=10\Rightarrow x^2+4=10\Rightarrow x^2=6\Rightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\end{matrix}\right.\)
a: f(-1/2)=1/4+4=17/4
f(5)=25+4=29
b: f(x)=10
nên \(x^2=6\)
hay \(x\in\left\{\sqrt{6};-\sqrt{6}\right\}\)