Question

$$x + \frac { 1 } { x } = 2 x ^ { 2024 } + \frac { 1 } { x ^ { 2014 } } = \cdots \cdots \cdots$$

Answer

$$h=((If*x+1/x)/(2*x^2014)-1/2)/(e*t*n*x^10)$$

Solution


Regroup terms.
\[Ifx+\frac{1}{x}=2ethn{x}^{2024}+1\times {x}^{2014}\]
Simplify  \(1\times {x}^{2014}\)  to  \({x}^{2014}\).
\[Ifx+\frac{1}{x}=2ethn{x}^{2024}+{x}^{2014}\]
Factor out the common term \({x}^{2014}\).
\[Ifx+\frac{1}{x}={x}^{2014}(2ethn{x}^{10}+1)\]
Divide both sides by \({x}^{2014}\).
\[\frac{Ifx+\frac{1}{x}}{{x}^{2014}}=2ethn{x}^{10}+1\]
Subtract \(1\) from both sides.
\[\frac{Ifx+\frac{1}{x}}{{x}^{2014}}-1=2ethn{x}^{10}\]
Divide both sides by \(2\).
\[\frac{\frac{Ifx+\frac{1}{x}}{{x}^{2014}}-1}{2}=ethn{x}^{10}\]
Simplify  \(\frac{\frac{Ifx+\frac{1}{x}}{{x}^{2014}}-1}{2}\)  to  \(\frac{\frac{Ifx+\frac{1}{x}}{{x}^{2014}}}{2}-\frac{1}{2}\).
\[\frac{\frac{Ifx+\frac{1}{x}}{{x}^{2014}}}{2}-\frac{1}{2}=ethn{x}^{10}\]
Simplify  \(\frac{\frac{Ifx+\frac{1}{x}}{{x}^{2014}}}{2}\)  to  \(\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}\).
\[\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}=ethn{x}^{10}\]
Divide both sides by \(e\).
\[\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{e}=thn{x}^{10}\]
Divide both sides by \(t\).
\[\frac{\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{e}}{t}=hn{x}^{10}\]
Simplify  \(\frac{\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{e}}{t}\)  to  \(\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{et}\).
\[\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{et}=hn{x}^{10}\]
Divide both sides by \(n\).
\[\frac{\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{et}}{n}=h{x}^{10}\]
Simplify  \(\frac{\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{et}}{n}\)  to  \(\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{etn}\).
\[\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{etn}=h{x}^{10}\]
Divide both sides by \({x}^{10}\).
\[\frac{\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{etn}}{{x}^{10}}=h\]
Simplify  \(\frac{\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{etn}}{{x}^{10}}\)  to  \(\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{etn{x}^{10}}\).
\[\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{etn{x}^{10}}=h\]
Switch sides.
\[h=\frac{\frac{Ifx+\frac{1}{x}}{2{x}^{2014}}-\frac{1}{2}}{etn{x}^{10}}\]