Question

مخرج کسر (الف) را گویا کنید. حاصل عبارت (ب) را بدست آورید:

الف) \frac{8}{\sqrt{5} + \sqrt{3}} =

ب) \frac{1}{\sqrt3{x} - 1} - \frac{1}{x - 1} =

Explanation:

🆕 New Concept Discovered: Rationalizing Denominators
Eliminating radicals from the bottom of a fraction.

Step 1: Rationalize part (a)

To eliminate the square roots from the denominator of \(\frac{8}{\sqrt{5} + \sqrt{3}}\), we multiply both the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{5} - \sqrt{3}\).

$$ \frac{8}{\sqrt{5} + \sqrt{3}} \cdot \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{8(\sqrt{5} - \sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2} $$

Simplify the denominator using the difference of squares formula \((a-b)(a+b) = a^2 - b^2\):

$$ (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2 $$

Now, simplify the entire fraction:

$$ \frac{8(\sqrt{5} - \sqrt{3})}{2} = 4(\sqrt{5} - \sqrt{3}) = 4\sqrt{5} - 4\sqrt{3} $$

Step 2: Simplify part (b)

We need to find the value of the expression:

$$ \frac{1}{\sqrt[3]{x} - 1} - \frac{1}{x - 1} $$

First, recall the algebraic identity for the difference of cubes:

$$ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $$

If we let \(a = \sqrt[3]{x}\) and \(b = 1\), we get:

$$ x - 1 = (\sqrt[3]{x} - 1)(\sqrt[3]{x^2} + \sqrt[3]{x} + 1) $$

This means the common denominator for the two fractions is \(x - 1\). We can rewrite the first fraction by multiplying its numerator and denominator by \(\sqrt[3]{x^2} + \sqrt[3]{x} + 1\):

$$ \frac{1}{\sqrt[3]{x} - 1} = \frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}{x - 1} $$

Now, subtract the second fraction:

$$ \frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}{x - 1} - \frac{1}{x - 1} = \frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1 - 1}{x - 1} $$

$$ = \frac{\sqrt[3]{x^2} + \sqrt[3]{x}}{x - 1} $$

Answer:

Part (a):

$$ 4\sqrt{5} - 4\sqrt{3} $$

Part (b):

$$ \frac{\sqrt[3]{x^2} + \sqrt[3]{x}}{x - 1} $$