Câu hỏi của huynhthithucquyen

Question

a) 1 + $\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}$b) 1 - $\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{9}}}$c) -3 + $\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{3}}}}$

⭐Hannie⭐ · Accepted Answer

$a,1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}\
=1+\dfrac{1}{1+\dfrac{1}{\dfrac{3}{2}}}\
=1+\dfrac{1}{1+\dfrac{2}{3}}\
=1+\dfrac{1}{\dfrac{5}{3}}\
=1+\dfrac{3}{5}\
=\dfrac{8}{5}$$b,1-\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{9}}}\
=1-\dfrac{1}{1-\dfrac{1}{\dfrac{8}{9}}}\
=1-\dfrac{1}{1-\dfrac{9}{8}}\
=1-\dfrac{1}{-\dfrac{1}{8}}\=1-\left(-8\right)\
=9$$c,-3+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{3}}}}\
=-3+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{\dfrac{4}{3}}}}\
=-3+\dfrac{1}{1+\dfrac{1}{3+\dfrac{3}{4}}}\
=-3+\dfrac{1}{1+\dfrac{1}{\dfrac{15}{4}}}\
=-3+\dfrac{1}{1+\dfrac{4}{15}}\
=-3+\dfrac{1}{\dfrac{19}{15}}\
=-3+\dfrac{15}{19}\
=-\dfrac{57}{19}+\dfrac{15}{19}\
=-\dfrac{42}{19}$Câu trả lời: $a,1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}\
=1+\dfrac{1}{1+\dfrac{1}{\dfrac{3}{2}}}\
=1+\dfrac{1}{1+\dfrac{2}{3}}\
=1+\dfrac{1}{\dfrac{5}{3}}\
=1+\dfrac{3}{5}\
=\dfrac{8}{5}$$b,1-\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{9}}}\
=1-\dfrac{1}{1-\dfrac{1}{\dfrac{8}{9}}}\
=1-\dfrac{1}{1-\dfrac{9}{8}}\
=1-\dfrac{1}{-\dfrac{1}{8}}\=1-\left(-8\right)\
=9$$c,-3+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{3}}}}\
=-3+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{\dfrac{4}{3}}}}\
=-3+\dfrac{1}{1+\dfrac{1}{3+\dfrac{3}{4}}}\
=-3+\dfrac{1}{1+\dfrac{1}{\dfrac{15}{4}}}\
=-3+\dfrac{1}{1+\dfrac{4}{15}}\
=-3+\dfrac{1}{\dfrac{19}{15}}\
=-3+\dfrac{15}{19}\
=-\dfrac{57}{19}+\dfrac{15}{19}\
=-\dfrac{42}{19}$

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