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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