Solve the following quadratic equation by factorization, the root is: -1 and 1
$\dfrac{x-1}{2x+1}+\dfrac{2x+1}{x-1}=\dfrac{5}{2},x\ne -\dfrac{1}{2},1$
a) True
b) False

Hint: Simplify the given relation by taking LCM of the denominator and hence cross-multiplying the equation. Use the given definition for $x\Rightarrow x\ne \dfrac{-1}{2},1$. Use the relation ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab, {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to solve the expression.

Complete step-by-step answer:
Given quadratic equation in the problem is
$\dfrac{x-1}{2x+1}+\dfrac{2x+1}{x-1}=\dfrac{5}{2}x\ne \dfrac{-1}{2},1.................\left( i \right)$
As, we know general equation of any quadratic equation is given as
$a{{x}^{2}}+bx+c=0$
But in the question the given equation is not in the general form. It means we need to simplify the given equation to get in the form of a general quadratic equation. So, we have
$\dfrac{x-1}{2x+1}+\dfrac{2x+1}{x-1}=\dfrac{5}{2}$
Now, we can take LCM of denominator to simplify the expression. So, we get
$\begin{align}
  & \dfrac{x-1}{2x+1}+\dfrac{2x+1}{x-1}=\dfrac{5}{2} \\
 & \dfrac{\left( x-1 \right)\left( x-1 \right)+\left( 2x+1 \right)\left( 2x+1 \right)}{\left( 2x+1 \right)\left( x-1 \right)}=\dfrac{5}{2} \\
 & \Rightarrow \dfrac{{{\left( x-1 \right)}^{2}}+{{\left( 2x+1 \right)}^{2}}}{\left( 2x+1 \right)\left( x-1 \right)}=\dfrac{5}{2} \\
\end{align}$
Now, use the algebraic identity of ${{\left( a-b \right)}^{2}},{{\left( a+b \right)}^{2}}$ to solve the above expression. So, we know identity as
\[\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align}\]
Hence, we get expression as
$\begin{align}
  & \dfrac{{{x}^{2}}+1-2x+4{{x}^{2}}+1+4x}{\left( 2x+1 \right)\left( x-1 \right)}=\dfrac{5}{2} \\
 & \Rightarrow \dfrac{5{{x}^{2}}+2x+2}{\left( 2x+1 \right)\left( x-1 \right)}=\dfrac{5}{2} \\
\end{align}$
Now, multiply the terms given in the denominator as well. So, we get
$\begin{align}
  & \dfrac{5{{x}^{2}}+2x+2}{2{{x}^{2}}-2x+x-1}=\dfrac{5}{2} \\
 & \Rightarrow \dfrac{5{{x}^{2}}+2x+2}{2{{x}^{2}}-x-1}=\dfrac{5}{2} \\
\end{align}$
On cross multiplying the above expression, we get the expression as
$\begin{align}
  & 2\left( 5{{x}^{2}}+2x+2 \right)=5\left( 2{{x}^{2}}-x-1 \right) \\
 & \Rightarrow 10{{x}^{2}}+4x+4=10{{x}^{2}}-5x-5 \\
 & 9x+9=0 \\
 & x=\dfrac{-9}{9}=-1 \\
 & \Rightarrow x=-1 \\
\end{align}$
Now, we get that the given equation is not a quadratic equation, it’s a linear equation and the value of ‘x’ is -1 only. Now, it is given in the problem root of the equation is ‘1’ as well, which is not possible as we cannot put ‘1’ to the given expression because term $\dfrac{2x+1}{x-1}$ will be infinite is x = 1. So, x = 1 is not in the domain of the question. It means the given statement is false.

Note: One may think that the given question is wrong as it is representing a linear equation x = -1, but it is not. We can write x = -1 as $0.{{x}^{2}}+1.x+1=0$ as well. Son don’t confuse the term “quadratic equation”.
Another direct approach for getting whether the statement is false or not, would be that as the roots of the equation are given as -1 and 1 and we can notice that 1 is not lying in its domain. So, we don’t need to solve the expression as well for the given problem.