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Cho a, b, x là những số dương. Đơn giản các biểu thức sau:
a) \(A = {{\rm{[}}{{2a + {{(ab)}^{{1 \over 2}}}} \over {3a}}{\rm{]}}^{ – 1}}{\rm{[}}{{{a^{{3 \over 2}}} – {b^{{3 \over 2}}}} \over {a – {{(ab)}^{{1 \over 2}}}}} – {{a – b} \over {\sqrt a + \sqrt b }}{\rm{]}}\)
b) \(B = {({{\sqrt a + \sqrt x } \over {\sqrt {a + x} }} – {{\sqrt {a + x} } \over {\sqrt a + \sqrt x }})^{ – 2}} – {({{\sqrt a – \sqrt x } \over {\sqrt {a + x} }} – {{\sqrt {a + x} } \over {\sqrt a – \sqrt x }})^{ – 2}}\)
c) \(C = \sqrt {{{16}^{{1 \over {{{\log }_7}4}}}} + {{81}^{{1 \over {{{\log }_6}9}}}} + 15} \)
d) \(D = {49^{1 – {{\log }_7}2}} + {5^{ – {{\log }_5}4}}\)
Hướng dẫn làm bài
Do a, b, x là những số dương nên ta có:
a) \({A_1} = {{\rm{[}}{{2a + {{(ab)}^{{1 \over 2}}}} \over {3a}}{\rm{]}}^{ – 1}} = {{3a} \over {2a + {{(ab)}^{{1 \over 2}}}}} = {{3{a^{{1 \over 2}}}} \over {2{a^{{1 \over 2}}} + {b^{{1 \over 2}}}}}\)
\({A_2} = \left[ {{{{a^{{3 \over 2}}} – {b^{{3 \over 2}}}} \over {a – {{(ab)}^{{1 \over 2}}}}}} \right. \left. {{{a – b} \over {\sqrt a + \sqrt b }}} \right]\)
\(= {{({a^{{1 \over 2}}} – {b^{{1 \over 2}}})(a + {{(ab)}^{{1 \over 2}}} + b)} \over {{a^{{1 \over 2}}}({a^{{1 \over 2}}} – {b^{{1 \over 2}}})}} – ({a^{{1 \over 2}}} – {b^{{1 \over 2}}})\)
\( = {{a + {{(ab)}^{{1 \over 2}}} + b – {a^{{1 \over 2}}}({a^{{1 \over 2}}} – {b^{{1 \over 2}}})} \over {{a^{{1 \over 2}}}}} = {{2{a^{{1 \over 2}}}{b^{{1 \over 2}}} + b} \over {{a^{{1 \over 2}}}}} = {{{b^{{1 \over 2}}}(2{a^{{1 \over 2}}} + {b^{{1 \over 2}}})} \over {{a^{{1 \over 2}}}}}\)
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Vậy \(A = {A_1}.{A_2} = {{3{a^{{1 \over 2}}}} \over {2{a^{{1 \over 2}}} + {b^{{1 \over 2}}}}}.{{{b^{{1 \over 2}}}(2{a^{{1 \over 2}}} + {b^{{1 \over 2}}})} \over {{a^{{1 \over 2}}}}} = 3\sqrt b \)
b) \({B_1} = {({{\sqrt a + \sqrt x } \over {\sqrt {a + x} }} – {{\sqrt {a + x} } \over {\sqrt a + \sqrt x }})^{ – 2}} = {{(a + x){{(\sqrt a + \sqrt x )}^2}} \over {4ax}}\)
\({B_2} = {({{\sqrt a – \sqrt x } \over {\sqrt {a + x} }} – {{\sqrt {a + x} } \over {\sqrt a – \sqrt x }})^{ – 2}} = {{(a + x){{(\sqrt a – \sqrt x )}^2}} \over {4ax}}\)
Vậy \(B = {B_1} – {B_2} = {{a + x} \over {\sqrt {ax} }}\)
c) Ta có \({16^{{1 \over {{{\log }_7}4}}}} = {4^{2{{\log }_4}7}} = 49;{81^{{1 \over {{{\log }_6}9}}}} = 36\)
\(\Rightarrow C = \sqrt {49 + 36 + 15} = 10\)
d) Ta có \({49^{1 – {{\log }_7}2}} = {{49} \over {{{49}^{{{\log }_7}2}}}} = {{49} \over 4};{5^{ – {{\log }_5}4}} = {1 \over 4}\)
\(\Rightarrow D = {{49} \over 4} + {1 \over 4} = {{25} \over 2}\).
Mục lục môn Toán 12(SBT)