Certainly! Let's solve the expression step-by-step:
Given the expression:
[tex]\[
\frac{\sqrt{m+n} - \sqrt{m-n}}{\sqrt{m+n} + \sqrt{m-n}}
\][/tex]
To simplify this expression, we can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\(\sqrt{m+n} + \sqrt{m-n}\)[/tex] is [tex]\(\sqrt{m+n} - \sqrt{m-n}\)[/tex].
We perform the multiplication as follows:
[tex]\[
\frac{(\sqrt{m+n} - \sqrt{m-n})(\sqrt{m+n} - \sqrt{m-n})}{(\sqrt{m+n} + \sqrt{m-n})(\sqrt{m+n} - \sqrt{m-n})}
\][/tex]
First, let's deal with the denominator:
[tex]\[
(\sqrt{m+n} + \sqrt{m-n})(\sqrt{m+n} - \sqrt{m-n})
\][/tex]
Using the difference of squares formula, [tex]\((a + b)(a - b) = a^2 - b^2\)[/tex]:
[tex]\[
(\sqrt{m+n})^2 - (\sqrt{m-n})^2
\][/tex]
[tex]\[
= (m+n) - (m-n)
\][/tex]
[tex]\[
= m + n - m + n
\][/tex]
[tex]\[
= 2n
\][/tex]
Next, let's simplify the numerator:
[tex]\[
(\sqrt{m+n} - \sqrt{m-n})^2
\][/tex]
Using the expansion of [tex]\((a - b)^2 = a^2 - 2ab + b^2\)[/tex]:
[tex]\[
= (\sqrt{m+n})^2 - 2(\sqrt{m+n})(\sqrt{m-n}) + (\sqrt{m-n})^2
\][/tex]
[tex]\[
= (m + n) - 2\sqrt{(m+n)(m-n)} + (m - n)
\][/tex]
[tex]\[
= m + n + m - n - 2\sqrt{(m+n)(m-n)}
\][/tex]
[tex]\[
= 2m - 2\sqrt{(m+n)(m-n)}
\][/tex]
Now, putting it all together:
[tex]\[
\frac{2m - 2\sqrt{(m+n)(m-n)}}{2n}
\][/tex]
We can factor out the 2 in the numerator:
[tex]\[
\frac{2(m - \sqrt{(m+n)(m-n)})}{2n}
\][/tex]
[tex]\[
= \frac{m - \sqrt{(m+n)(m-n)}}{n}
\][/tex]
So, the simplified form of the expression is:
[tex]\[
\frac{m - \sqrt{(m+n)(m-n)}}{n}
\][/tex]
