Bài 8 trang 78 SGK Đại số và Giải tích 12 Nâng cao

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Đơn giản biểu thức:

LG 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:

${{\sqrt a - \sqrt b } \over {\root 4 \of a - \root 4 \of b }} - {{\sqrt a + \root 4 \of {ab} } \over {\root 4 \of a + \root 4 \of b }}$

$= {{\left( {\root 4 \of a + \root 4 \of b } \right)\left( {\root 4 \of a - \root 4 \of b } \right)} \over {\root 4 \of a - \root 4 \of b }} - {{\root 4 \of a \left( {\root 4 \of a + \root 4 \of b } \right)} \over {\root 4 \of a + \root 4 \of b }}$

$= \root 4 \of a + \root 4 \of b - \root 4 \of a = \root 4 \of b$

LG 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:

${{a - b} \over {\root 3 \of a - \root 3 \of b }} - {{a + b} \over {\root 3 \of a + \root 3 \of b }}$

$= {{{{\left( {\root 3 \of a } \right)}^3} - {{\left( {\root 3 \of b } \right)}^3}} \over {\root 3 \of a - \root 3 \of b }} - {{{{\left( {\root 3 \of a } \right)}^3} + {{\left( {\root 3 \of b } \right)}^3}} \over {\root 3 \of a + \root 3 \of b }}$

$= \frac{{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)\left( {\sqrt[3]{{{a^2}}} + \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)}}{{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)}}$ $- \frac{{\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)\left( {\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)}}{{\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)}}$

$= (\root 3 \of {{a^2}} + \root 3 \of {ab} + \root 3 \of {{b^2}})$ $- \left( {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} } \right)$

$= 2\root 3 \of {ab}$

LG 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:

$\left( {{{a + b} \over {\root 3 \of a + \root 3 \of b }} - \root 3 \of {ab} } \right):{\left( {\root 3 \of a - \root 3 \of b } \right)^2}$

$= (\frac{{\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)\left( {\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right)}}{{\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)}}- \root 3 \of {ab}):$ $:{\left( {\sqrt[3]{a} - \sqrt[3]{b}} \right)^2}$

$= \left( {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} - \root 3 \of {ab} } \right):$ $:{\left( {\root 3 \of a - \root 3 \of b } \right)^2}$

$= \left( {\root 3 \of {{a^2}} - 2\root 3 \of {ab} + \root 3 \of {{b^2}} } \right):{\left( {\root 3 \of a - \root 3 \of b } \right)^2}$

$= {\left( {\root 3 \of a - \root 3 \of b } \right)^2}:{\left( {\root 3 \of a - \root 3 \of b } \right)^2} = 1$

LG 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:

${{a - 1} \over {{a^{{3 \over 4}}} + {a^{{1 \over 2}}}}}.{{\sqrt a + \root 4 \of a } \over {\sqrt a + 1}}.{a^{{1 \over 4}}} + 1.$

$= {{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)} \over {\sqrt[4]{{{a^3}}} + \sqrt a }}.{{\root 4 \of a \left( {\root 4 \of a + 1} \right)} \over {\left( {\sqrt a + 1} \right)}}.\root 4 \of a + 1$

$= \frac{{\left( {\sqrt a + 1} \right)\left( {\sqrt[4]{a} + 1} \right)\left( {\sqrt[4]{a} - 1} \right)}}{{\sqrt a \left( {\sqrt[4]{a} + 1} \right)}}.\frac{{\sqrt[4]{a}\left( {\sqrt[4]{a} + 1} \right)}}{{\sqrt a + 1}}.\sqrt[4]{a} + 1$

$= \frac{{\left( {\sqrt[4]{a} + 1} \right)\left( {\sqrt[4]{a} - 1} \right).{{\left( {\sqrt[4]{a}} \right)}^2}}}{{\sqrt a }} + 1$

$= \sqrt a - 1 + 1 = \sqrt a$ .

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