we know that
A rational number is any number that can be expressed as a ratio of two integers.
case A) [tex]6\sqrt{3}- \sqrt{108}[/tex]
we know that
[tex]108=2^{2}3^{3}[/tex]
so
[tex]\sqrt{108}=\sqrt{2^{2}3^{3}}=6\sqrt{3}[/tex]
substitute in the expression
[tex]6\sqrt{3}- 6\sqrt{3}=0[/tex]
The number [tex]0[/tex] is a rational number since it can be written as [tex]0/2[/tex]
case B) [tex]\pi \sqrt{9}[/tex]
we know that
[tex]\pi \sqrt{9}=3 \pi[/tex]
The number [tex]3 \pi[/tex] is not a rational number because it can't be expressed as a relationship of two integers
case C) [tex]\sqrt{49}+ \sqrt{5}[/tex]
we know that
[tex]\sqrt{49}+ \sqrt{5}=7+\sqrt{5}[/tex]
The number [tex]7+\sqrt{5}[/tex] is not a rational number because it can't be expressed as a relationship of two integers
case D) [tex](5\sqrt{3})(4\sqrt{3})[/tex]
we know that
[tex](5\sqrt{3})(4\sqrt{3})=(5)(4)(\sqrt{3})(\sqrt{3})=(20)(3)=60[/tex]
The number [tex]60[/tex] is a rational number since it can be written as [tex]60/1[/tex]
case E) [tex]\frac{1}{3} +\frac{4}{5}[/tex]
we know that
[tex]\frac{1}{3} +\frac{4}{5}=\frac{5*1+3*4}{5*3}= \frac{17}{15}[/tex]
The number [tex]\frac{17}{15}[/tex] is a rational number since it can be expressed as a ratio of two integers
therefore
the answer is
[tex]6\sqrt{3}- \sqrt{108}[/tex]
[tex](5\sqrt{3})(4\sqrt{3})[/tex]
[tex](5\sqrt{3})(4\sqrt{3})[/tex]
