Answer:
[tex]\Large\boxed{\sf \bf \ \ \dfrac{1}{3x+52} \ \ }[/tex]
Step-by-step explanation:
Hello, please consider the following.
We need to do something with that, right !?
[tex]\dfrac{\left(\dfrac{1}{x^2+51x+50\right)}}{\left(\dfrac{2}{x+50}+\dfrac{1}{x+1}\right)}[/tex]
What can we say from [tex]x^2+51x+50[/tex] ?
The sum of the zeroes is -51=(-1)+(-50) and the product is 50 = (-1) x (-50), so we can factorise. Let's do it !
[tex]x^2+51x+50=x^2+50x+x+50=x(x+1)+50(x+1)=(x+1)(x+50)[/tex]
That's a pretty cool first result !
Now, let's play with the denominator.
[tex]\dfrac{2}{x+50}+\dfrac{1}{x+1}\\\\\text{*** We put on the same denominator which is (x+1)(x+50) ***}\\\\=\dfrac{2(x+1)}{(x+50)(x+1)}+\dfrac{x+50}{(x+1)(x+50)}\\\\=\dfrac{2(x+1)+x+50}{(x+50)(x+1)}\\\\=\dfrac{2x+2+x+50}{(x+50)(x+1)}\\\\=\dfrac{3x+52}{(x+50)(x+1)}\\[/tex]
We are almost there.
Let's combine all these results together !
[tex]\dfrac{\left(\dfrac{1}{x^2+51x+50\right)}}{\left(\dfrac{2}{x+50}+\dfrac{1}{x+1}\right)}\\\\\\=\dfrac{\left(\dfrac{1}{(x+1)(x+50)\right)}}{\left(\dfrac{3x+52}{(x+50)(x+1)}\right)}}\\\\\\=\large\boxed{\dfrac{1}{3x+52}}[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
