Bài tập 8 trang 78 SGK Toán 12 NC

Bài tập 8 trang 78 SGK Toán 12 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{{\sqrt a  - \sqrt b }}{{\sqrt[4]{a} - \sqrt[4]{b}}} - \frac{{\sqrt a  + \sqrt[4]{{ab}}}}{{\sqrt[4]{a} + \sqrt[4]{b}}}\\
 = \frac{{\left( {\sqrt[4]{a} + \sqrt[4]{b}} \right)\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)}}{{\sqrt[4]{a} - \sqrt[4]{b}}}\\
 - \frac{{\sqrt[4]{a}\left( {\sqrt[4]{a} + \sqrt[4]{b}} \right)}}{{\sqrt[4]{a} + \sqrt[4]{b}}}
\end{array}\\
{ = \sqrt[4]{a} + \sqrt[4]{b} - \sqrt[4]{a} = \sqrt[4]{b}}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{{a - b}}{{\sqrt[3]{a} - \sqrt[3]{b}}} - \frac{{a + b}}{{\sqrt[3]{a} + \sqrt[3]{b}}}\\
 = \frac{{{{\left( {\sqrt[3]{a}} \right)}^3} - {{\left( {\sqrt[3]{b}} \right)}^3}}}{{\sqrt[3]{a} - \sqrt[3]{b}}} - \frac{{{{\left( {\sqrt[3]{a}} \right)}^3} + {{\left( {\sqrt[3]{b}} \right)}^3}}}{{\sqrt[3]{a} + \sqrt[3]{b}}}
\end{array}\\
\begin{array}{l}
 = \sqrt[3]{{{a^2}}} + \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}} - \left( {\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)\\
 = 2\sqrt[3]{{ab}}
\end{array}
\end{array}\)

c)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\left( {\frac{{a + b}}{{\sqrt[3]{a} + \sqrt[3]{b}}} - \sqrt[3]{{ab}}} \right):{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)^2}\\
 = \left( {\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}} - \sqrt[3]{{ab}}} \right):{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)^2}
\end{array}\\
{ = \left( {\sqrt[3]{{{a^2}}} - 2\sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right):{{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)}^2}}\\
{ = {{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)}^2}:{{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)}^2} = 1}
\end{array}\)

d)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{{a - 1}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}}}.\frac{{\sqrt a  + \sqrt[4]{a}}}{{\sqrt a  + 1}}.{a^{\frac{1}{4}}} + 1\\
 = \frac{{\left( {\sqrt a  + 1} \right)\left( {\sqrt a  - 1} \right)}}{{\sqrt a .\left( {\sqrt[4]{a} + 1} \right)}}.\frac{{\sqrt[4]{a}\left( {\sqrt[4]{a} + 1} \right)}}{{\left( {\sqrt a  + 1} \right)}}.\sqrt[4]{a} + 1
\end{array}\\
{ = \sqrt a  - 1 + 1 = \sqrt a }
\end{array}\)

-- Mod Toán 12