$81\cdot\left(\dfrac{12-\dfrac{12}{7}-\dfrac{12}{289}-\dfrac{12}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}:\dfrac{5+\dfrac{5}{13}+\dfrac5{169}

$81\cdot\left(\dfrac{12-\dfrac{12}{7}-\dfrac{12}{289}-\dfrac{12}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}:\dfrac{5+\dfrac{5}{13}+\dfrac5{169}+\dfrac{5}{91}}{6+\dfrac{6}{13}+\dfrac{6}{169}+\dfrac{6}{91}}\right)\cdot\dfrac{\dfrac{5}{8}+\dfrac{5}{29}-\dfrac{5}{n}}{27\cdot\left(\dfrac{54}{8}+\dfrac{108}{58}+\dfrac{162^2}{3n^2}\right)}$

0 bình luận về “$81\cdot\left(\dfrac{12-\dfrac{12}{7}-\dfrac{12}{289}-\dfrac{12}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}:\dfrac{5+\dfrac{5}{13}+\dfrac5{169}”

  1. Đáp án:

    `1`

    Giải thích các bước giải:

    $81\cdot\left(\dfrac{12-\dfrac{12}{7}-\dfrac{12}{289}-\dfrac{12}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}:\dfrac{5+\dfrac{5}{13}+\dfrac5{169}+\dfrac{5}{91}}{6+\dfrac{6}{13}+\dfrac{6}{169}+\dfrac{6}{91}}\right)\cdot\dfrac{\dfrac{5}{8}+\dfrac{5}{29}-\dfrac{5}{n}}{27\cdot\left(\dfrac{54}{8}+\dfrac{108}{58}+\dfrac{162^2}{3n^2}\right)}$

    $81\cdot\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{\ 4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac1{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{5\left(\dfrac{1}{8}+\dfrac{1}{29}-\dfrac{1}{n}\right)}{27\cdot\left(\dfrac{54}{8}+\dfrac{54}{29}-\dfrac{54}{n}\right)}$

    $=81\cdot\left(\dfrac{12}{4}:\dfrac{5}{6}\right)\cdot\dfrac{5\left(\dfrac{1}{8}+\dfrac{1}{29}-\dfrac{1}{n}\right)}{27.54\left(\dfrac{1}{8}+\dfrac{1}{29}-\dfrac{1}{n}\right)}$

    $=81\cdot\left(3:\dfrac{5}{6}\right)\cdot\dfrac{5}{27.54}$

    $=81\cdot\dfrac{18}{5}\cdot\dfrac{5}{1458}$

    $=\dfrac{1458}{5}\cdot\dfrac{5}{1458}$

    $=1$

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