Question

1. \\(\frac{1}{\\_} = \frac{1}{x^2 + 2x} + \frac{x - 1}{x}\\) (note: there seems to be a typo or incomplete part in the original left - hand side of the equation, but we extract the available text as much as possible)

Explanation:

Step1: Simplify the denominator

First, factor the denominator \(x^{2}+2x\) as \(x(x + 2)\). So the equation becomes \(\frac{1}{x(x + 2)}=\frac{1}{x(x + 2)}+\frac{x - 1}{x}\) (assuming the left - hand side was a typo and should be \(\frac{1}{x^{2}+2x}\) as the right - hand side has two terms. Wait, maybe the left - hand side is \(\frac{1}{x}\)? Wait, the original equation: Let's re - examine. If the equation is \(\frac{1}{x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\) (maybe a misprint in the left - hand side). Let's assume the correct equation is \(\frac{1}{x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\) or \(\frac{1}{x^{2}+2x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\) (but the second one would lead to \(\frac{x - 1}{x}=0\) which is \(x = 1\), but let's check the original. Wait, maybe the left - hand side is \(\frac{1}{x}\), so:

Starting over, let's assume the equation is \(\frac{1}{x}=\frac{1}{x(x + 2)}+\frac{x - 1}{x}\) (factoring \(x^{2}+2x=x(x + 2)\)).

Step2: Multiply through by \(x(x + 2)\) (the least common denominator) to eliminate fractions

Multiply each term by \(x(x + 2)\):

\(x(x + 2)\times\frac{1}{x}=x(x + 2)\times\frac{1}{x(x + 2)}+x(x + 2)\times\frac{x - 1}{x}\)

Simplify each term:

First term: \(x + 2\)

Second term: \(1\)

Third term: \((x + 2)(x - 1)=x^{2}+2x-x - 2=x^{2}+x - 2\)

So the equation becomes:

\(x + 2=1+x^{2}+x - 2\)

Step3: Simplify the equation

Subtract \(x\) from both sides:

\(x + 2-x=1+x^{2}+x - 2-x\)

\(2=x^{2}-1\)

Step4: Solve for \(x\)

Add \(1\) to both sides: \(x^{2}=3\)

Take square roots: \(x=\pm\sqrt{3}\)

Wait, but let's check the original equation again. Maybe the left - hand side was \(\frac{1}{x^{2}+2x}\), so the equation is \(\frac{1}{x(x + 2)}=\frac{1}{x(x + 2)}+\frac{x - 1}{x}\)

Subtract \(\frac{1}{x(x + 2)}\) from both sides:

\(0=\frac{x - 1}{x}\)

Which implies \(x-1 = 0\) (since \(x eq0\) as it makes the denominator zero), so \(x = 1\)

Wait, there must be a misprint in the original problem. Let's assume the correct equation is \(\frac{1}{x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\)

Subtract \(\frac{x - 1}{x}\) from both sides:

\(\frac{1}{x}-\frac{x - 1}{x}=\frac{1}{x^{2}+2x}\)

\(\frac{1-(x - 1)}{x}=\frac{1}{x(x + 2)}\)

\(\frac{1 - x + 1}{x}=\frac{1}{x(x + 2)}\)

\(\frac{2 - x}{x}=\frac{1}{x(x + 2)}\)

Multiply both sides by \(x(x + 2)\) ( \(x eq0,x eq - 2\)):

\((2 - x)(x + 2)=1\)

\(4-x^{2}=1\)

\(x^{2}=3\)

\(x=\pm\sqrt{3}\)

But let's go back to the original problem's possible intention. If the equation is \(\frac{1}{x^{2}+2x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\), then subtracting \(\frac{1}{x^{2}+2x}\) from both sides gives \(\frac{x - 1}{x}=0\), so \(x-1 = 0\) ( \(x eq0\)), so \(x = 1\)

Let's check \(x = 1\) in the original (assumed) equation \(\frac{1}{1^{2}+2\times1}=\frac{1}{1^{2}+2\times1}+\frac{1 - 1}{1}\), which is \(\frac{1}{3}=\frac{1}{3}+0\), which holds.

If we take the equation as \(\frac{1}{x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\), then:

\(\frac{1}{x}-\frac{x - 1}{x}=\frac{1}{x^{2}+2x}\)

\(\frac{1-(x - 1)}{x}=\frac{1}{x(x + 2)}\)

\(\frac{2 - x}{x}=\frac{1}{x(x + 2)}\)

Multiply both sides by \(x(x + 2)\) ( \(x eq0,x eq - 2\)):

\((2 - x)(x + 2)=1\)

\(4-x^{2}=1\)

\(x^{2}=3\)

\(x=\pm\sqrt{3}\)

Checking \(x=\sqrt{3}\):

Left - hand side: \(\frac{1}{\sqrt{3}}\)

Right - hand side: \(\frac{1}{3 + 2\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}}\)

Simplify \(\frac{1}{3 + 2\sqrt{3}}=\frac{3 - 2\sqrt{3}}{(3 + 2\sqrt{3})(3 - 2\sqrt{3})}=\frac{3 - 2\sqrt{3}}{9-12}=\frac{3 - 2\sqrt{3}}{-3}=-1+\frac{2\sqrt{3}}{3}\)

\(\frac{\sqrt{3}-1}{\sqrt{3}} = 1-\frac{1}{\sqrt{3}}\)

Sum of right - hand side terms: \(-1+\frac{2\sqrt{3}}{3}+1-\frac{1}{\sqrt{3}}=\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{3} eq\frac{1}{\sqrt{3}}\) (wait, \(\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)), so it does hold.

But the most probable intended equation is when we have \(\frac{1}{x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\), leading to \(x=\pm\sqrt{3}\), or \(\frac{1}{x^{2}+2x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\) leading to \(x = 1\)

Wait, maybe the original equation was \(\frac{1}{x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\), and after subtracting \(\frac{x - 1}{x}\) from both sides, we get \(0=\frac{1}{x^{2}+2x}\), which has no solution, which can't be.

Let's re - express the original equation as it is written: \(\frac{1}{x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\) (assuming the left - hand side is \(\frac{1}{x}\))

Subtract \(\frac{x - 1}{x}\) from both sides:

\(\frac{1-(x - 1)}{x}=\frac{1}{x^{2}+2x}\)

\(\frac{2 - x}{x}=\frac{1}{x(x + 2)}\)

Multiply both sides by \(x(x + 2)\):

\((2 - x)(x + 2)=1\)

\(4-x^{2}=1\)

\(x^{2}=3\)

\(x=\pm\sqrt{3}\)

Answer:

If the equation is \(\frac{1}{x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\), the solutions are \(x = \sqrt{3}\) and \(x=-\sqrt{3}\). If the equation is \(\frac{1}{x^{2}+2x}=\frac{1}{x^{2}+2x}+\frac{x - 1}{x}\), the solution is \(x = 1\)