#### Answer

The required solution is False

#### Work Step by Step

We have the given algebraic expression:
$\frac{{{x}^{2}}-25}{x-5}=x-5$
We know that for an algebraic expression, a rational expression is an expression which can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator $q\ne 0$.
By solving the left-hand side of the given expression, we have:
$\frac{{{x}^{2}}-25}{x-5}$
And factorize ${{x}^{2}}-25$ using the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
$\begin{align}
& {{x}^{2}}-25={{\left( x \right)}^{2}}-{{\left( 5 \right)}^{2}} \\
& =\left( x+5 \right)\left( x-5 \right)
\end{align}$
So, the left-hand side of the given expression becomes:
$\frac{{{x}^{2}}-25}{x-5}=\frac{\left( x+5 \right)\left( x-5 \right)}{\left( x-5 \right)}$
And solve further to obtain the solution:
$\begin{align}
& \frac{{{x}^{2}}-25}{x-5}=\frac{\left( x+5 \right)\left( x-5 \right)}{\left( x-5 \right)} \\
& =x+5
\end{align}$
Hence, $\frac{{{x}^{2}}-25}{x-5}=$ $x+5$.