Tìm n biết:
\(\frac{{{8^7} + {8^7} + {8^7} + {8^7}}}{{{3^7} + {3^7} + {3^7}}}:\frac{{{2^7} + {2^7}}}{{{6^7} + {6^7} + {6^7} + {6^7} + {6^7} + {6^7}}} = {2^n}\)
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A.
24
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B.
23
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C.
25
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D.
8
Rút gọn vế trái
Nếu am = an ( a khác 0, a khác 1) thì m = n
$\frac{{{8}^{7}}+{{8}^{7}}+{{8}^{7}}+{{8}^{7}}}{{{3}^{7}}+{{3}^{7}}+{{3}^{7}}}:\frac{{{2}^{7}}+{{2}^{7}}}{{{6}^{7}}+{{6}^{7}}+{{6}^{7}}+{{6}^{7}}+{{6}^{7}}+{{6}^{7}}}={{2}^{n}}$
$\frac{{{4.8}^{7}}}{{{3.3}^{7}}}:\frac{{{2.2}^{7}}}{{{6.6}^{7}}}={{2}^{n}}$
$\frac{{{2}^{2}}.{{\left( {{2}^{3}} \right)}^{7}}}{{{3}^{8}}}:\frac{{{2}^{8}}}{{{6}^{8}}}={{2}^{n}}$
$\frac{{{2}^{2}}{{.2}^{3.7}}}{{{3}^{8}}}.\frac{{{6}^{8}}}{{{2}^{8}}}={{2}^{n}}$
$\frac{{{2}^{2}}{{.2}^{21}}{{.6}^{8}}}{{{(3.2)}^{8}}}={{2}^{n}}$
$\frac{{{2}^{2+21}}{{.6}^{8}}}{{{6}^{8}}}={{2}^{n}}$
${{2}^{23}}={{2}^{n}}$
$23=n$
Vậy n = 23
Đáp án : B









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