Perform the following operations s and prove closure. Show your work.

Answer:1. [tex]\frac{x}{x+3}+\frac{x+2}{x+5}[/tex] = [tex]\frac{2x^2+10x+6}{(x+3)(x+5)}\\[/tex]2. [tex]\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}[/tex] = [tex]\frac{1}{(x+2)(x-4)}[/tex]3. [tex]\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}[/tex] = [tex]\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}[/tex]4. [tex]\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}[/tex] = [tex]\frac{1}{(x-2)(x-4)}[/tex]Step-by-step explanation:1. [tex]\frac{x}{x+3}+\frac{x+2}{x+5}[/tex]Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)[tex]=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\[/tex]Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.2. [tex]\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}[/tex]Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)Putting factors[tex]=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}[/tex]Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.3. [tex]\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}[/tex]Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)Putting factors[tex]\frac{2}{(x+3)(x-3)}-\frac{3x}{(x+3)(x-2)}[/tex]Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)[tex]\frac{2(x-2)-3x(x+3)(x-3)}{(x+3)(x-3)(x-2)}[/tex][tex]=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}[/tex]Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.4. [tex]\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}[/tex]Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)[tex]\frac{x+4}{(x-2)(x+3)}\div\frac{(x-4)(x+4)}{x+3}[/tex]Converting ÷ sign into multiplication we will take reciprocal of the second term[tex]=\frac{x+4}{(x-2)(x+3)}*\frac{x+3}{(x-4)(x+4)}\\=\frac{1}{(x-2)(x-4)}[/tex]Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.