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Câu hỏi: Đơn giản biểu thức:
Lời giải chi tiết:
\({{\sqrt a - \sqrt b } \over {\root 4 \of a - \root 4 \of b }} - {{\sqrt a + \root 4 \of {ab} } \over {\root 4 \of a + \root 4 \of b }}\)
\(= {{\left( {\root 4 \of a + \root 4 \of b } \right)\left({\root 4 \of a - \root 4 \of b } \right)} \over {\root 4 \of a - \root 4 \of b }} - {{\root 4 \of a \left({\root 4 \of a + \root 4 \of b } \right)} \over {\root 4 \of a + \root 4 \of b }}\)
\(= \root 4 \of a + \root 4 \of b - \root 4 \of a = \root 4 \of b \)
Lời giải chi tiết:
\({{a - b} \over {\root 3 \of a - \root 3 \of b }} - {{a + b} \over {\root 3 \of a + \root 3 \of b }} \)
\(= {{{{\left( {\root 3 \of a } \right)}^3} - {{\left({\root 3 \of b } \right)}^3}} \over {\root 3 \of a - \root 3 \of b }} - {{{{\left({\root 3 \of a } \right)}^3} + {{\left({\root 3 \of b } \right)}^3}} \over {\root 3 \of a + \root 3 \of b }}\)
\(= \frac{{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)\left({\sqrt[3]{{{a^2}}} + \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)}}{{\left({\sqrt[3]{a} - \sqrt[3]{b}} \right)}}\) \(- \frac{{\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)\left({\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)}}{{\left({\sqrt[3]{a} + \sqrt[3]{b}} \right)}}\)
\(= (\root 3 \of {{a^2}} + \root 3 \of {ab} + \root 3 \of {{b^2}}) \) \(- \left( {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} } \right) \)
\(= 2\root 3 \of {ab} \)
Lời giải chi tiết:
\(\left( {{{a + b} \over {\root 3 \of a + \root 3 \of b }} - \root 3 \of {ab} } \right):{\left({\root 3 \of a - \root 3 \of b } \right)^2}\)
\(= (\frac{{\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)\left({\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)}}{{\left({\sqrt[3]{a} + \sqrt[3]{b}} \right)}}- \root 3 \of {ab}):\) \(:{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)^2}\)
\(= \left( {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} - \root 3 \of {ab} } \right):\) \(:{\left( {\root 3 \of a - \root 3 \of b } \right)^2}\)
\(= \left( {\root 3 \of {{a^2}} - 2\root 3 \of {ab} + \root 3 \of {{b^2}} } \right):{\left({\root 3 \of a - \root 3 \of b } \right)^2} \)
\(= {\left( {\root 3 \of a - \root 3 \of b } \right)^2}:{\left({\root 3 \of a - \root 3 \of b } \right)^2} = 1\)
Lời giải chi tiết:
\({{a - 1} \over {{a^{{3 \over 4}}} + {a^{{1 \over 2}}}}}.{{\sqrt a + \root 4 \of a } \over {\sqrt a + 1}}.{a^{{1 \over 4}}} + 1. \)
\(= {{\left( {\sqrt a + 1} \right)\left({\sqrt a - 1} \right)} \over {\sqrt[4]{{{a^3}}} + \sqrt a }}.{{\root 4 \of a \left({\root 4 \of a + 1} \right)} \over {\left({\sqrt a + 1} \right)}}.\root 4 \of a + 1\)
\(= \frac{{\left( {\sqrt a + 1} \right)\left({\sqrt[4]{a} + 1} \right)\left({\sqrt[4]{a} - 1} \right)}}{{\sqrt a \left({\sqrt[4]{a} + 1} \right)}}.\frac{{\sqrt[4]{a}\left({\sqrt[4]{a} + 1} \right)}}{{\sqrt a + 1}}.\sqrt[4]{a} + 1\)
\(= \frac{{\left( {\sqrt[4]{a} + 1} \right)\left({\sqrt[4]{a} - 1} \right).{{\left({\sqrt[4]{a}} \right)}^2}}}{{\sqrt a }} + 1\)
\(= \sqrt a - 1 + 1 = \sqrt a \).
Cách khác:
$\begin{array}{l}
\frac{{a - 1}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}}}.\frac{{\sqrt a + \sqrt[4]{a}}}{{\sqrt a + 1}}.{a^{\frac{1}{4}}} + 1\\
= \frac{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}}}.\frac{{\left( {{a^{\frac{1}{2}}} + {a^{\frac{1}{4}}}} \right){a^{\frac{1}{4}}}}}{{\sqrt a + 1}} + 1\\
= \frac{{\left( {\sqrt a - 1} \right)\left( {{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}} \right)}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}}} + 1
\end{array}$
Câu a
\({{\sqrt a - \sqrt b } \over {\root 4 \of a - \root 4 \of b }} - {{\sqrt a + \root 4 \of {ab} } \over {\root 4 \of a + \root 4 \of b }}\)Lời giải chi tiết:
\({{\sqrt a - \sqrt b } \over {\root 4 \of a - \root 4 \of b }} - {{\sqrt a + \root 4 \of {ab} } \over {\root 4 \of a + \root 4 \of b }}\)
\(= {{\left( {\root 4 \of a + \root 4 \of b } \right)\left({\root 4 \of a - \root 4 \of b } \right)} \over {\root 4 \of a - \root 4 \of b }} - {{\root 4 \of a \left({\root 4 \of a + \root 4 \of b } \right)} \over {\root 4 \of a + \root 4 \of b }}\)
\(= \root 4 \of a + \root 4 \of b - \root 4 \of a = \root 4 \of b \)
Câu b
\({{a - b} \over {\root 3 \of a - \root 3 \of b }} - {{a + b} \over {\root 3 \of a + \root 3 \of b }}\)Lời giải chi tiết:
\({{a - b} \over {\root 3 \of a - \root 3 \of b }} - {{a + b} \over {\root 3 \of a + \root 3 \of b }} \)
\(= {{{{\left( {\root 3 \of a } \right)}^3} - {{\left({\root 3 \of b } \right)}^3}} \over {\root 3 \of a - \root 3 \of b }} - {{{{\left({\root 3 \of a } \right)}^3} + {{\left({\root 3 \of b } \right)}^3}} \over {\root 3 \of a + \root 3 \of b }}\)
\(= \frac{{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)\left({\sqrt[3]{{{a^2}}} + \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)}}{{\left({\sqrt[3]{a} - \sqrt[3]{b}} \right)}}\) \(- \frac{{\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)\left({\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)}}{{\left({\sqrt[3]{a} + \sqrt[3]{b}} \right)}}\)
\(= (\root 3 \of {{a^2}} + \root 3 \of {ab} + \root 3 \of {{b^2}}) \) \(- \left( {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} } \right) \)
\(= 2\root 3 \of {ab} \)
Câu c
\(\left( {{{a + b} \over {\root 3 \of a + \root 3 \of b }} - \root 3 \of {ab} } \right):{\left({\root 3 \of a - \root 3 \of b } \right)^2};\)Lời giải chi tiết:
\(\left( {{{a + b} \over {\root 3 \of a + \root 3 \of b }} - \root 3 \of {ab} } \right):{\left({\root 3 \of a - \root 3 \of b } \right)^2}\)
\(= (\frac{{\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)\left({\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)}}{{\left({\sqrt[3]{a} + \sqrt[3]{b}} \right)}}- \root 3 \of {ab}):\) \(:{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)^2}\)
\(= \left( {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} - \root 3 \of {ab} } \right):\) \(:{\left( {\root 3 \of a - \root 3 \of b } \right)^2}\)
\(= \left( {\root 3 \of {{a^2}} - 2\root 3 \of {ab} + \root 3 \of {{b^2}} } \right):{\left({\root 3 \of a - \root 3 \of b } \right)^2} \)
\(= {\left( {\root 3 \of a - \root 3 \of b } \right)^2}:{\left({\root 3 \of a - \root 3 \of b } \right)^2} = 1\)
Câu d
\({{a - 1} \over {{a^{{3 \over 4}}} + {a^{{1 \over 2}}}}}.{{\sqrt a + \root 4 \of a } \over {\sqrt a + 1}}.{a^{{1 \over 4}}} + 1.\)Lời giải chi tiết:
\({{a - 1} \over {{a^{{3 \over 4}}} + {a^{{1 \over 2}}}}}.{{\sqrt a + \root 4 \of a } \over {\sqrt a + 1}}.{a^{{1 \over 4}}} + 1. \)
\(= {{\left( {\sqrt a + 1} \right)\left({\sqrt a - 1} \right)} \over {\sqrt[4]{{{a^3}}} + \sqrt a }}.{{\root 4 \of a \left({\root 4 \of a + 1} \right)} \over {\left({\sqrt a + 1} \right)}}.\root 4 \of a + 1\)
\(= \frac{{\left( {\sqrt a + 1} \right)\left({\sqrt[4]{a} + 1} \right)\left({\sqrt[4]{a} - 1} \right)}}{{\sqrt a \left({\sqrt[4]{a} + 1} \right)}}.\frac{{\sqrt[4]{a}\left({\sqrt[4]{a} + 1} \right)}}{{\sqrt a + 1}}.\sqrt[4]{a} + 1\)
\(= \frac{{\left( {\sqrt[4]{a} + 1} \right)\left({\sqrt[4]{a} - 1} \right).{{\left({\sqrt[4]{a}} \right)}^2}}}{{\sqrt a }} + 1\)
\(= \sqrt a - 1 + 1 = \sqrt a \).
Cách khác:
$\begin{array}{l}
\frac{{a - 1}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}}}.\frac{{\sqrt a + \sqrt[4]{a}}}{{\sqrt a + 1}}.{a^{\frac{1}{4}}} + 1\\
= \frac{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}}}.\frac{{\left( {{a^{\frac{1}{2}}} + {a^{\frac{1}{4}}}} \right){a^{\frac{1}{4}}}}}{{\sqrt a + 1}} + 1\\
= \frac{{\left( {\sqrt a - 1} \right)\left( {{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}} \right)}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}}} + 1
\end{array}$
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