We have
[tex]\begin{array}{l}f(x) = \sqrt{\dfrac{|x+3| + x}{x+2} - 1}\\\\f(x) = \sqrt{\dfrac{|x+3| + x - x - 2}{x+2}}\\\\f(x) = \sqrt{\dfrac{|x+3|- 2}{x+2}}\end{array}[/tex]
Now,
[tex]{x + 2 \not = 0\implies x \in \mathbb R - \{-2\}}[/tex]
This is equation (1.)
Also,
[tex]{\dfrac{|x+3| - 2}{x+2}\ge 0\implies x \in [-5,-2)\cup[-1,\infty)}[/tex]
This is equation (2.)
Using equation (1.) and (2.)
x ∈ [-5, -2) ∪ [-1, ∞)
B is correct bro.
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Verified answer
We have
[tex]\begin{array}{l}f(x) = \sqrt{\dfrac{|x+3| + x}{x+2} - 1}\\\\f(x) = \sqrt{\dfrac{|x+3| + x - x - 2}{x+2}}\\\\f(x) = \sqrt{\dfrac{|x+3|- 2}{x+2}}\end{array}[/tex]
Now,
[tex]{x + 2 \not = 0\implies x \in \mathbb R - \{-2\}}[/tex]
This is equation (1.)
Also,
[tex]{\dfrac{|x+3| - 2}{x+2}\ge 0\implies x \in [-5,-2)\cup[-1,\infty)}[/tex]
This is equation (2.)
Using equation (1.) and (2.)
x ∈ [-5, -2) ∪ [-1, ∞)
B is correct bro.