Question

PLEASE help me with this question! This is URGENT!

Answer 1

Answer:

$\frac{1}{3x+52}$

Step-by-step explanation:

We begin with the expression

$\frac{\frac{1}{x^2+51x+50} }{\frac{2}{x+50}+\frac{1}{x+1} }$

First, we can split this into the product of two fractions

$\frac{1}{x^2+51x+50} *(\frac{1}{\frac{2}{x+50}+\frac{1}{x+1}} )$

Now, we can factor the denominator of the first fraction.

$\frac{1}{(x+1)(x+50)} *(\frac{1}{\frac{2}{x+50}+\frac{1}{x+1}} )$

Next, we need to create a common denominator for the second fraction

$\frac{2}{x+50}+\frac{1}{x+1}\\\\\frac{2}{x+50}*\frac{x+1}{x+1} +\frac{1}{x+1}*\frac{x+50}{x+50} \\\\\frac{2x+2}{(x+1)(x+50)} + \frac{x+50}{(x+1)(x+50)}\\\\ \frac{3x+52}{(x+1)(x+50)}$

Now to return this portion to the denominator of the second fraction.

$\frac{1}{(x+1)(x+50)} *(\frac{1}{\frac{3x+52}{(x+1)(x+50)} } )$

Dividing by a fraction is the same as multiplying as the reciprocal of the fraction, so the second fraction becomes

$\frac{1}{(x+1)(x+50)} *\frac{(x+1)(x+50)}{3x+52}$

Now to reduce this expression by canceling out $(x+1)(x+5)$

$\frac{(x+1)(x+50)}{(x+1)(x+50)(3x+52)}\\\\\frac{1}{3x+52}$