$Domainoftherealvaluedfunction\sqrt{25-9x^2}+\sqrt{x^2+x+1}is$

$\begin{array}{l}f\left(x\right)=\sqrt{25-9x^2}+\sqrt{x^2+x+1}isdefined\\ iff25-9x^2\ge 0and{x}^2+x+1\ge 0\\ \Rightarrow {\left(3x\right)}^2-5^2\le 0, x^2+2\left(\frac{1}{2}\right)x+\frac{1}{4}-\frac{1}{4}+1\ge 0\\ \Rightarrow (3x+5)(3x-5)\le 0, (x+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.)+\frac{3}{4}\ge 0\forall x\in R\\ \therefore x\in [\frac{-5}{3},\frac{5}{3}]\\ \therefore Domain=[\frac{-5}{3},\frac{5}{3}]\end{array}$