Question

Question #c28d1

Answer 1

Assuming that I reformatted the question correctly
`(a+(a/(1+a/(a+1))))=color(red)((3a+1)/(2a+1)a`

Explanation:

`(a+(a/(1+a/(a+1))))`

`color(white)("XXX")=(a+(a/((a+1+a)/a)))`

`color(white)("XXX")=(a+a/((2a+1)/a))`

`color(white)("XXX")=(a+a * a/(2a+1))`

`color(white)("XXX")=(a+(a^2)/(2a+1))`

`color(white)("XXX")=(a(2a+1)+a^2)/(2a+1)`

`color(white)("XXX")=(2a^2+a+a^2)/(2a+1)`

`color(white)("XXX")=(3a^2+a)/(2a+1)`

`color(white)("XXX")=(3a+1)/(2a+1) * a`

Answer 2

`(a+(a/(1+a/(a+1))))=(a(3a+2))/(2a+1)`

Explanation:

`(a+(a/(1+a/(a+1))))`

Let's start with just the fraction.

Multiply the fraction by `color(blue)((a+1)/(a+1))`

`a/(1+a/(a+1))*color(blue)(a+1)/color(blue)(a+1)=(a(a+1))/((a+1)(1+a/(a+1)))=(a(a+1))/((a+1)+a(a+1)/(a+1)))=(a^2+a)/((a+1)+a)=color(red)((a^2+a)/(2a+1))`

Now multiply the first term by `color(blue)((2a+1)/(2a+1))` to get a common denominator

`(a+color(red)((a^2+a)/(2a+1)))=(a*(color(blue)(2a+1)/color(blue)(2a+1))+color(red)((a^2+a)/(2a+1)))=(2a^2+a)/(2a+1)+color(red)((a^2+a)/(2a+1))=(2a^2+a+a^2+a)/(2a+1)=(3a^2+2a)/(2a+1)=(a(3a+2))/(2a+1)`