Question

$$\left. \begin{array} { l } { e ( x ) = \frac { 2 + 1 } { 2 x - 1 } , z \neq 1 } \\ { = } \end{array} \right.$$

Answer

$$f=(z^2*p*m-1)/x,(2*e^2*nFCFCF*n*t*h)/x$$

Solution


Regroup terms.
\[fx=zzpm-1,2nethenFCFCF\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[fx={z}^{2}pm-1,2nethenFCFCF\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[fx={z}^{2}pm-1,2n{e}^{2}thnFCFCF\]
Regroup terms.
\[fx={z}^{2}pm-1,2{e}^{2}nFCFCFnth\]
Break down the problem into these 2 equations.
\[fx={z}^{2}pm-1\]
\[fx=2{e}^{2}nFCFCFnth\]
Solve the 1st equation: \(fx={z}^{2}pm-1\).
\[f=\frac{{z}^{2}pm-1}{x}\]
Solve the 2nd equation: \(fx=2{e}^{2}nFCFCFnth\).
\[f=\frac{2{e}^{2}nFCFCFnth}{x}\]
Collect all solutions.
\[f=\frac{{z}^{2}pm-1}{x},\frac{2{e}^{2}nFCFCFnth}{x}\]