Expression Evaluation: Simplifying Fractions with Square Roots

The problem asks for the approximate value of the sum \(\frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}}\). We can simplify this expression by rationalizing the denominators of each fraction.

Rationalizing the First Term

Consider the first term: \(\frac{1}{1+\sqrt{2}}\).

• Multiply the numerator and the denominator by the conjugate of the denominator, which is \(1-\sqrt{2}\): \(\frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}\)
• Apply the difference of squares formula \((a+b)(a-b) = a^2 - b^2\) to the denominator: \(\frac{1-\sqrt{2}}{1^2 - (\sqrt{2})^2} = \frac{1-\sqrt{2}}{1-2} = \frac{1-\sqrt{2}}{-1}\)
• Simplify the fraction: \(\sqrt{2} - 1\)

Rationalizing the Second Term

Consider the second term: \(\frac{1}{\sqrt{2}+\sqrt{3}}\). It's easier to write the denominator as \(\sqrt{3}+\sqrt{2}\).

• Multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{3}-\sqrt{2}\): \(\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
• Apply the difference of squares formula to the denominator: \(\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3}-\sqrt{2}}{3-2} = \frac{\sqrt{3}-\sqrt{2}}{1}\)
• Simplify the fraction: \(\sqrt{3} - \sqrt{2}\)

Summing the Simplified Terms

Now, add the results from the two rationalized terms:

• Sum = \((\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2})\)
• Combine like terms. The \(\sqrt{2}\) and \(-\sqrt{2}\) cancel out: \(\text{Sum} = \sqrt{3} - 1\)

Approximating the Value

To find the value closest to the options, we approximate \(\sqrt{3}\).

• We know that \(\sqrt{3} \approx 1.732\).
• Substitute this value into the simplified sum: \(\text{Value} \approx 1.732 - 1 = 0.732\)

Comparing $0.732$ to the given options, it is closest to $0.73$.