Question

$$x ^ { 4 } + \frac { 1 } { x ^ { 4 } } x - \frac { 1 } { x } = 5$$

Answer

$$f=(5-e*w*h*n*x^-3+1/x)/x^4$$

Solution


Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[f{x}^{4}+\frac{1\times whenx}{{x}^{4}}-\frac{1}{x}=5\]
Simplify  \(1\times whenx\)  to  \(whnxe\).
\[f{x}^{4}+\frac{whnxe}{{x}^{4}}-\frac{1}{x}=5\]
Use Quotient Rule: \(\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}\).
\[f{x}^{4}+whn{x}^{1-4}e-\frac{1}{x}=5\]
Simplify  \(1-4\)  to  \(-3\).
\[f{x}^{4}+whn{x}^{-3}e-\frac{1}{x}=5\]
Regroup terms.
\[f{x}^{4}+ewhn{x}^{-3}-\frac{1}{x}=5\]
Subtract \(ewhn{x}^{-3}\) from both sides.
\[f{x}^{4}-\frac{1}{x}=5-ewhn{x}^{-3}\]
Add \(\frac{1}{x}\) to both sides.
\[f{x}^{4}=5-ewhn{x}^{-3}+\frac{1}{x}\]
Divide both sides by \({x}^{4}\).
\[f=\frac{5-ewhn{x}^{-3}+\frac{1}{x}}{{x}^{4}}\]