Question

(3 7/18 - 1 1/9)*36/41+(8 - 3 4/9): 41/18
помогите(

Answer 1

Ответ:

$\left(3\dfrac{7}{18} -1\dfrac{1}{9}\right )\cdot\dfrac{36}{41} +\left(8-3\dfrac{4}{9} \right):\dfrac{41}{18}=4$

Пошаговое объяснение:

$\left(3\dfrac{7}{18} -1\dfrac{1}{9}\right )\cdot\dfrac{36}{41} +\left(8-3\dfrac{4}{9} \right):\dfrac{41}{18}$

Установим порядок действий.

$\left(3\dfrac{7}{18} \stackrel{(1)}{\;-}1\dfrac{1}{9}\right )\stackrel{(3)}{\;\cdot}\dfrac{36}{41} \stackrel{(5)}{\;+}\left(8\stackrel{(2)}{\;-}3\dfrac{4}{9} \right)\stackrel{(4)}{\;:}\dfrac{41}{18}$

$1) 3\dfrac{7}{18} -1\dfrac{1}{9}=(3-1)+\bigg( {\overset{1/}{\big{}}} \dfrac{7}{18}-{\overset{2/}{\big{}}} \dfrac{1}{9} \frac{}{} \bigg)=2+\dfrac{7-2}{18} =2\dfrac{5}{18} ;$

$2)2\dfrac{5}{18} \cdot\dfrac{36}{41} =\dfrac{41}{18} \cdot\dfrac{36}{41} =\dfrac{41\cdot36}{18\cdot41} =\dfrac{36}{18} =2;$

$3)8-3\dfrac{4}{9}=7\dfrac{9}{9} -3\dfrac{4}{9}=(7-3)+\dfrac{9-4}{9} =4\dfrac{5}{9} ;$

$4)4\dfrac{5}{9} :\dfrac{41}{18} =\dfrac{41}{9} \cdot\dfrac{18}{41} =\dfrac{41\cdot18}{9\cdot41} =\dfrac{18}{9} =2;$

$5)2+2=4$

Значит,

$\left(3\dfrac{7}{18} -1\dfrac{1}{9}\right )\cdot\dfrac{36}{41} +\left(8-3\dfrac{4}{9} \right):\dfrac{41}{18}=4$