Question

1. \\(\frac{5}{x^{3}+5x^{2}} = \frac{4}{x + 5} + \frac{1}{x^{2}}\\)

Explanation:

Step1: Factor the denominator on the left

Factor \(x^{3}+5x^{2}\) as \(x^{2}(x + 5)\), so the equation becomes \(\frac{5}{x^{2}(x + 5)}=\frac{4}{x + 5}+\frac{1}{x^{2}}\).

Step2: Find the least common denominator (LCD)

The LCD of \(x^{2}(x + 5)\), \(x + 5\), and \(x^{2}\) is \(x^{2}(x + 5)\). Multiply each term by \(x^{2}(x + 5)\) to eliminate the fractions:
\[

$$\begin{align*} \frac{5}{x^{2}(x + 5)}\times x^{2}(x + 5)&=\frac{4}{x + 5}\times x^{2}(x + 5)+\frac{1}{x^{2}}\times x^{2}(x + 5)\\ 5&=4x^{2}+(x + 5) \end{align*}$$

\]

Step3: Simplify the equation

Simplify the right - hand side: \(5 = 4x^{2}+x + 5\). Subtract 5 from both sides: \(4x^{2}+x+5 - 5=0\), which simplifies to \(4x^{2}+x = 0\).

Step4: Factor the quadratic equation

Factor out an \(x\) from \(4x^{2}+x\): \(x(4x + 1)=0\).

Step5: Solve for \(x\)

Set each factor equal to zero:

• If \(x = 0\), the original equation has denominators \(x^{3}+5x^{2}=0\), \(x + 5 = 5\) (but \(x = 0\) makes the first and third denominators zero), so \(x = 0\) is an extraneous solution.
• If \(4x+1 = 0\), then \(4x=-1\), and \(x=-\frac{1}{4}\).

Step6: Check the solution

Substitute \(x = -\frac{1}{4}\) into the original equation:

• Left - hand side: \(\frac{5}{(-\frac{1}{4})^{3}+5\times(-\frac{1}{4})^{2}}=\frac{5}{-\frac{1}{64}+\frac{5}{16}}=\frac{5}{\frac{- 1 + 20}{64}}=\frac{5}{\frac{19}{64}}=\frac{320}{19}\)
• Right - hand side: \(\frac{4}{-\frac{1}{4}+5}+\frac{1}{(-\frac{1}{4})^{2}}=\frac{4}{\frac{19}{4}}+16=\frac{16}{19}+16=\frac{16 + 304}{19}=\frac{320}{19}\)

So \(x = -\frac{1}{4}\) is a valid solution.

Answer:

\(x=-\frac{1}{4}\)