## Precalculus (6th Edition) Blitzer

The factor of the given equation $\frac{\sqrt{x}-\frac{1}{4\sqrt{x}}}{\sqrt{x}}$ is $\frac{4x-1}{4x}$ .
Consider the expression: $\frac{\sqrt{x}-\frac{1}{4\sqrt{x}}}{\sqrt{x}}$ Multiply and divide the first term of the numerator $4\sqrt{x}$ $\frac{\sqrt{x}-\frac{1}{4\sqrt{x}}}{\sqrt{x}}=\frac{\frac{\sqrt{x}\times 4\sqrt{x}}{4\sqrt{x}}-\frac{1}{4\sqrt{x}}}{\sqrt{x}}$ Apply the radical rule: $\sqrt{a}\cdot \sqrt{a}=a$ $\frac{\frac{\sqrt{x}\times 4\sqrt{x}}{4\sqrt{x}}-\frac{1}{4\sqrt{x}}}{\sqrt{x}}=\frac{\frac{4x-1}{4\sqrt{x}}}{\sqrt{x}}$ Apply the fraction rule: $\frac{\frac{b}{c}}{a}=\frac{b}{c\cdot a}$ $\frac{4x-1}{4\sqrt{x}\sqrt{x}}$ Apply the radical rule: $\sqrt{a}\cdot \sqrt{a}=a$ $\frac{\frac{4x-1}{4\sqrt{x}}}{\sqrt{x}}=\frac{4x-1}{4x}$ The simplified form of the expression $\frac{\sqrt{x}-\frac{1}{4\sqrt{x}}}{\sqrt{x}}$ is $\frac{4x-1}{4x}$ .