We are asked to rationalize the numerator for each of the three given expressions. (i) $\frac{\sqrt{5+h}-3}{h}$ (ii) $\frac{\sqrt{3}+\sqrt{5}}{7}$ (iii) $\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}+\sqrt{x+h}}$

1. Problem Description

We are asked to rationalize the numerator for each of the three given expressions.

(i) $\frac{\sqrt{5+h}-3}{h}$

(ii) $\frac{\sqrt{3}+\sqrt{5}}{7}$

(iii) $\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}+\sqrt{x+h}}$

2. Solution Steps

(i) Rationalize the numerator of $\frac{\sqrt{5+h}-3}{h}$.

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{5+h}+3$.

$\frac{\sqrt{5+h}-3}{h} \cdot \frac{\sqrt{5+h}+3}{\sqrt{5+h}+3} = \frac{(5+h) - 9}{h(\sqrt{5+h}+3)} = \frac{h-4}{h(\sqrt{5+h}+3)}$

(ii) Rationalize the numerator of $\frac{\sqrt{3}+\sqrt{5}}{7}$.

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{3}-\sqrt{5}$.

$\frac{\sqrt{3}+\sqrt{5}}{7} \cdot \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}} = \frac{3-5}{7(\sqrt{3}-\sqrt{5})} = \frac{-2}{7(\sqrt{3}-\sqrt{5})}$

(iii) Rationalize the numerator of $\frac{\sqrt{x}-\sqrt{x+h}}{h(\sqrt{x}+\sqrt{x+h})}$.

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x}+\sqrt{x+h}$.

$\frac{\sqrt{x}-\sqrt{x+h}}{h(\sqrt{x}+\sqrt{x+h})} \cdot \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} = \frac{x - (x+h)}{h(\sqrt{x}+\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})} = \frac{-h}{h(\sqrt{x}+\sqrt{x+h})^2} = \frac{-1}{(\sqrt{x}+\sqrt{x+h})^2} = \frac{-1}{x + 2\sqrt{x(x+h)} + x+h} = \frac{-1}{2x+h+2\sqrt{x^2+xh}}$

3. Final Answer

(i) $\frac{h-4}{h(\sqrt{5+h}+3)}$

(ii) $\frac{-2}{7(\sqrt{3}-\sqrt{5})}$

(iii) $\frac{-1}{(\sqrt{x}+\sqrt{x+h})^2}$

or equivalently:

(i) $\frac{h-4}{h(\sqrt{5+h}+3)}$

(ii) $\frac{-2}{7(\sqrt{3}-\sqrt{5})}$

(iii) $\frac{-1}{2x+h+2\sqrt{x^2+xh}}$