Solving the Algebraic Equation

The problem asks us to find the value of a specific expression using a given algebraic equation. The equation is:

where \(m, a, b > 0\). We need to find the value of \(\frac{x^2+a b}{x^2+m^2 a b}\).

Let's simplify the given equation. We can cross-multiply:

Expand the terms in the numerators and denominators:

This expression looks symmetric. Let's move all terms to one side:

Let's use a substitution to simplify the structure. Let \(R = x^2 + ab\), \(S = (a+b)x\), \(P = x^2 + m^2 ab\), \(Q = m(a+b)x\). The equation can be written as:

Expand both sides:

Subtract \(RP - SQ\) from both sides:

Rearrange terms to get like terms on one side:

Substitute back the original expressions for R, Q, S, and P:

Since \(a, b > 0\), \(a+b \neq 0\). If we assume \(x \neq 0\), we can divide both sides by \((a+b)x\):

Rearrange the terms to solve for \(x^2\):

This equation implies either \(m-1 = 0\) or \(x^2 = mab\).

• If \(m-1 = 0\), then \(m=1\). The equation becomes \(0=0\). The expression \(\frac{x^2+ab}{x^2+m^2 ab}\) becomes \(\frac{x^2+ab}{x^2+ab} = 1\) (if denominator non-zero).
• If \(m \neq 1\), then we must have \(x^2 = mab\). Since \(m, a, b > 0\), \(mab > 0\), so real values for \(x\) exist.

The problem expects a unique value for the expression. Let's evaluate the expression using the relationship \(x^2 = mab\).

Evaluating the Expression \(\frac{x^2+a b}{x^2+m^2 a b}\)

Now substitute the value of \(x^2\) derived from the equation (\(x^2 = mab\)) into the expression \(\frac{x^2+a b}{x^2+m^2 a b}\):

Factor out common terms from the numerator and the denominator. In the numerator, \(ab\) is a common factor. In the denominator, \(mab\) is a common factor, or we can factor out \(ab\):

In the denominator, we can further factor out \(m\):

Since \(a, b > 0\) and \(m > 0\), \(ab \neq 0\) and \(m+1 \neq 0\). We can cancel the common factor \(ab(m+1)\) from the numerator and the denominator:

This is the value of the expression obtained when \(x^2=mab\).

If \(m=1\), \(x^2=ab\). The expression is \(\frac{ab+ab}{ab+1^2 ab} = \frac{2ab}{2ab} = 1\). Our result \(\frac{1}{m}\) becomes \(\frac{1}{1} = 1\), which is consistent.

Comparing with Options

Let's compare our derived value with the given options:

1. \(-\frac{1}{m^2} \)
2. \(\frac{1}{\mathrm{~m}^2}\)
3. \(\frac{2}{\mathrm{~m}}\)
4. \(\frac{1}{\mathrm{~m}}\)

Our calculated value \(\frac{1}{m}\) matches option 4.

Revision Table: Algebraic Simplification

Additional Information: Potential Solutions of the Equation

The equation \(x^2(m - 1) = mab(m - 1)\) derived from the original equation implies either \(m=1\) or \(x^2=mab\).

If \(m=1\), the original equation simplifies to \(1=1\) (for valid \(x\)), and the expression \(\frac{x^2+ab}{x^2+m^2ab}\) becomes \(\frac{x^2+ab}{x^2+ab}=1\). Our result \(\frac{1}{m}\) also gives \(\frac{1}{1}=1\).

If \(m \neq 1\), the equation implies \(x^2=mab\). Since \(m, a, b > 0\), \(mab > 0\), so \(x = \pm\sqrt{mab}\) are the real solutions for \(x\). For these values of \(x\), the expression evaluates to \(\frac{1}{m}\).

There was also the possibility in the derivation \(RQ=SP\) that \((a+b)x=0\). Since \(a,b>0\), \(a+b \neq 0\), so this implies \(x=0\). If \(x=0\), the original equation becomes \(\frac{ab}{m^2ab} = \frac{ab}{m^2ab}\), which is \(\frac{1}{m^2} = \frac{1}{m^2}\) (assuming \(ab \neq 0\)). So \(x=0\) is also a possible solution to the equation. However, if \(x=0\), the expression \(\frac{x^2+ab}{x^2+m^2ab}\) evaluates to \(\frac{0+ab}{0+m^2ab} = \frac{ab}{m^2ab} = \frac{1}{m^2}\).

The problem structure and the provided options suggest that the intended solution path leads to \(\frac{1}{m}\), corresponding to the \(x^2=mab\) case, or perhaps that \(\frac{1}{m}\) is the value that holds true for all valid \(x\) satisfying the equation simultaneously (though this is only true when \(m=1\)). Based on typical problems of this form, the relationship \(x^2=mab\) derived from simplifying the equation is usually the intended path to evaluate the expression.