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Sagot :
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]
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]
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