Question

URGENT! PLEASE help me! Full solutions please, and no nonsense answers.

Answer 1

`\Large\boxed{\sf \bf \ \ (1)/(3x+52) \ \ }`

Explanation:

Hello, please consider the following.

We need to do something with that, right !?

`(\left((1)/(x^2+51x+50\right)))/(\left((2)/(x+50)+(1)/(x+1)\right))`

What can we say from
`x^2+51x+50` ?

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 !

`x^2+51x+50=x^2+50x+x+50=x(x+1)+50(x+1)=(x+1)(x+50)`

That's a pretty cool first result !

Now, let's play with the denominator.

`(2)/(x+50)+(1)/(x+1)\\\\\text{*** We put on the same denominator which is (x+1)(x+50) ***}\\\\=(2(x+1))/((x+50)(x+1))+(x+50)/((x+1)(x+50))\\\\=(2(x+1)+x+50)/((x+50)(x+1))\\\\=(2x+2+x+50)/((x+50)(x+1))\\\\=(3x+52)/((x+50)(x+1))\\`

We are almost there.

Let's combine all these results together !

`(\left((1)/(x^2+51x+50\right)))/(\left((2)/(x+50)+(1)/(x+1)\right))\\\\\\=(\left((1)/((x+1)(x+50)\right)))/(\left((3x+52)/((x+50)(x+1))\right))}\\\\\\=\large\boxed{(1)/(3x+52)}`

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer 2

`(1)/(3x+52)`

Explanation:

Given

`((1)/(x^2+51x+50) )/((2)/(x+50)+(1)/(x+1) )`

=
`((1)/((x+50)(x+1)) )/((2(x+1)+x+50)/((x+50)(x+1)) )`

=
`(1)/((x+50)(x+1))` ×
`((x+50)(x+1))/(2x+2+x+50)` ← cancel (x + 50)(x + 1) on numerator/denominator

=
`(1)/(3x+52)`