Bài 11 trang 217 SBT giải tích 12

LG a

$A = {{\rm{[}}{{2a + {{(ab)}^{{1 \over 2}}}} \over {3a}}{\rm{]}}^{ - 1}}{\rm{[}}{{{a^{{3 \over 2}}} - {b^{{3 \over 2}}}} \over {a - {{(ab)}^{{1 \over 2}}}}} - {{a - b} \over {\sqrt a  + \sqrt b }}{\rm{]}}$

Lời giải chi tiết:

Do a, b, x là những số dương nên ta có:

$\displaystyle {A_1} = {{\rm{[}}{{2a + {{(ab)}^{{1 \over 2}}}} \over {3a}}{\rm{]}}^{ - 1}}$ $\displaystyle = {{3a} \over {2a + {{(ab)}^{{1 \over 2}}}}}$ $\displaystyle = {{3{a^{{1 \over 2}}}} \over {2{a^{{1 \over 2}}} + {b^{{1 \over 2}}}}}$

 $\displaystyle {A_2} = \left[ {{{{a^{{3 \over 2}}} - {b^{{3 \over 2}}}} \over {a - {{(ab)}^{{1 \over 2}}}}}} \right. - \left. {{{a - b} \over {\sqrt a  + \sqrt b }}} \right]$

$\displaystyle = {{({a^{{1 \over 2}}} - {b^{{1 \over 2}}})(a + {{(ab)}^{{1 \over 2}}} + b)} \over {{a^{{1 \over 2}}}({a^{{1 \over 2}}} - {b^{{1 \over 2}}})}} - ({a^{{1 \over 2}}} - {b^{{1 \over 2}}})$

$\displaystyle = {{a + {{(ab)}^{{1 \over 2}}} + b - {a^{{1 \over 2}}}({a^{{1 \over 2}}} - {b^{{1 \over 2}}})} \over {{a^{{1 \over 2}}}}}$ $\displaystyle  = {{2{a^{{1 \over 2}}}{b^{{1 \over 2}}} + b} \over {{a^{{1 \over 2}}}}}$

$\displaystyle = {{{b^{{1 \over 2}}}(2{a^{{1 \over 2}}} + {b^{{1 \over 2}}})} \over {{a^{{1 \over 2}}}}}$

Vậy $\displaystyle A = {A_1}.{A_2}$ $\displaystyle = {{3{a^{{1 \over 2}}}} \over {2{a^{{1 \over 2}}} + {b^{{1 \over 2}}}}}.{{{b^{{1 \over 2}}}(2{a^{{1 \over 2}}} + {b^{{1 \over 2}}})} \over {{a^{{1 \over 2}}}}}$ $\displaystyle = 3\sqrt b$

LG b

$D = {49^{1 - {{\log }_7}2}} + {5^{ - {{\log }_5}4}}$

Lời giải chi tiết:

Ta có  $\displaystyle {49^{1 - {{\log }_7}2}} = {{49} \over {{{49}^{{{\log }_7}2}}}} = {{49} \over 4};$ $\displaystyle  {5^{ - {{\log }_5}4}} = {1 \over 4}$

$\displaystyle \Rightarrow D = {{49} \over 4} + {1 \over 4} = {{25} \over 2}$.

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