91Ó°ÊÓ

In calculus, it is sometimes desirable to rationalize the numerator. To rationalize a numerator, we multiply the numerator and the denominator by the conjugate of the numerator. For example, $$\frac{6-\sqrt{2}}{4}=\frac{(6-\sqrt{2})(6+\sqrt{2})}{4(6+\sqrt{2})}=\frac{36-2}{4(6+\sqrt{2})}=\frac{34}{4(6+\sqrt{2})}=\frac{17}{2(6+\sqrt{2})}$$ Rationalize each numerator. Assume that all variables represent positive real numbers. $$ \frac{6-\sqrt{3}}{8} $$

Simplify. Assume that all variables represent positive real numbers. \(\sqrt{x} \cdot \sqrt[3]{x}\)

\(f(x)=\sqrt[3]{x}-3\)

Simplify. Assume that all variables represent positive real numbers. \(\sqrt[3]{y} \cdot \sqrt[4]{y}\)

In calculus, it is sometimes desirable to rationalize the numerator. To rationalize a numerator, we multiply the numerator and the denominator by the conjugate of the numerator. For example, $$\frac{6-\sqrt{2}}{4}=\frac{(6-\sqrt{2})(6+\sqrt{2})}{4(6+\sqrt{2})}=\frac{36-2}{4(6+\sqrt{2})}=\frac{34}{4(6+\sqrt{2})}=\frac{17}{2(6+\sqrt{2})}$$ Rationalize each numerator. Assume that all variables represent positive real numbers. $$ \frac{2 \sqrt{5}-3}{2} $$

\(f(x)=\sqrt[3]{x}+1\)

In calculus, it is sometimes desirable to rationalize the numerator. To rationalize a numerator, we multiply the numerator and the denominator by the conjugate of the numerator. For example, $$\frac{6-\sqrt{2}}{4}=\frac{(6-\sqrt{2})(6+\sqrt{2})}{4(6+\sqrt{2})}=\frac{36-2}{4(6+\sqrt{2})}=\frac{34}{4(6+\sqrt{2})}=\frac{17}{2(6+\sqrt{2})}$$ Rationalize each numerator. Assume that all variables represent positive real numbers. $$ \frac{2 \sqrt{x}-\sqrt{y}}{3 x} $$

\(f(x)=\sqrt[3]{x-3}\)

In calculus, it is sometimes desirable to rationalize the numerator. To rationalize a numerator, we multiply the numerator and the denominator by the conjugate of the numerator. For example, $$\frac{6-\sqrt{2}}{4}=\frac{(6-\sqrt{2})(6+\sqrt{2})}{4(6+\sqrt{2})}=\frac{36-2}{4(6+\sqrt{2})}=\frac{34}{4(6+\sqrt{2})}=\frac{17}{2(6+\sqrt{2})}$$ Rationalize each numerator. Assume that all variables represent positive real numbers. $$ \frac{\sqrt{p}-3 \sqrt{q}}{4 q} $$

\(\sqrt{19^{2}}\)