Bài 19 trang 82 SGK Đại số và Giải tích 12 Nâng cao

LG a

${a^{ - 2\sqrt 2 }}{\left( {{1 \over {{a^{ - \sqrt 2  - 1}}}}} \right)^{\sqrt 2  + 1}}$; 

Lời giải chi tiết:

${a^{ - 2\sqrt 2 }}{\left( {{1 \over {{a^{ - \sqrt 2  - 1}}}}} \right)^{\sqrt 2  + 1}}$

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

$= {a^{ - 2\sqrt 2 }}{\left( {{a^{\sqrt 2  + 1}}} \right)^{\sqrt 2  + 1}}$

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

$= {a^{ - 2\sqrt 2 }}{a^{3 + 2\sqrt 2 }}  = {a^{ - 2\sqrt 2  + 3 + 2\sqrt 2 }}$

$= {a^3}$

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LG b

${\left( {{{{a^{\sqrt 3 }}} \over {{b^{\sqrt 3  - 1}}}}} \right)^{\sqrt 3  + 1}}{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}};$

Lời giải chi tiết:

${\left( {{{{a^{\sqrt 3 }}} \over {{b^{\sqrt 3  - 1}}}}} \right)^{\sqrt 3  + 1}}{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}}$

$= \frac{{{{\left( {{a^{\sqrt 3 }}} \right)}^{\sqrt 3  + 1}}}}{{{{\left( {{b^{\sqrt 3  - 1}}} \right)}^{\sqrt 3  + 1}}}}.\frac{{{a^{ - 1 - \sqrt 3 }}}}{{{b^{ - 2}}}}$

$= \frac{{{a^{\sqrt 3 .\left( {\sqrt 3  + 1} \right)}}}}{{{b^{\left( {\sqrt 3  - 1} \right)\left( {\sqrt 3  + 1} \right)}}}}.\frac{{{a^{ - 1 - \sqrt 3 }}}}{{{b^{ - 2}}}}$

$= {{{a^{3 + \sqrt 3 }}} \over {{b^2}}}.{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}}$

$= \frac{{{a^{3 + \sqrt 3 }}.{a^{ - 1 - \sqrt 3 }}}}{{{b^2}.{b^{ - 2}}}} = \frac{{{a^{3 + \sqrt 3  - 1 - \sqrt 3 }}}}{{{b^{2 - 2}}}} = \frac{{{a^2}}}{{{b^0}}}= {a^2}$

LG c

${{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} + 1;$ 

Lời giải chi tiết:

${{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} + 1 = {{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }} + {{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}$

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

$= {{2{a^{2\sqrt 2 }} - 2{a^{\sqrt 2 }}{b^{\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}$

$= {{2{a^{\sqrt 2 }}\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}$

$= {{2{a^{\sqrt 2 }}} \over {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}}}$

LG d

$\sqrt {{{\left( {{x^\pi } + {y^\pi }} \right)}^2} - {{\left( {{4^{{1 \over \pi }}}xy} \right)}^\pi }} ;$

Lời giải chi tiết:

$\sqrt {{{\left( {{x^\pi } + {y^\pi }} \right)}^2} - {{\left( {{4^{{1 \over \pi }}}xy} \right)}^\pi }}$

$\begin{array}{l} = \sqrt {{{\left( {{x^\pi }} \right)}^2} + 2{x^\pi }{y^\pi } + {{\left( {{y^\pi }} \right)}^2} - {{\left( {{4^{\frac{1}{\pi }}}} \right)}^\pi }{x^\pi }{y^\pi }} \\ = \sqrt {{x^{2\pi }} + 2{x^\pi }{y^\pi } + {y^{2\pi }} - 4{x^\pi }{y^\pi }} \end{array}$

$= \sqrt {{x^{2\pi }} + {y^{2\pi }} - 2{x^\pi }{y^\pi }}$

$= \sqrt {{{\left( {{x^\pi } - {y^\pi }} \right)}^2}}$

$= \left| {{x^\pi } - {y^\pi }} \right|$.

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