Chapter A.10, Problem 87AYU

To determine

Express the expression $\frac{x}{{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+2{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ as a single quotient in which only positive exponent or radical appear.

Answer to Problem 87AYU

The required form of the given expression is $\frac{3x+2}{{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$.

Explanation of Solution

Given:

It is asked to determine the expression as a single quotient in which only positive exponents or radicals appear.

Calculation:

Consider $\frac{x}{{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+2{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$,

$\frac{x}{{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+2{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{x+2\left(1+x\right)}{{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{3x+2}{{\left(1+x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$