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\(A=\dfrac{\dfrac{3}{11}+\dfrac{3}{3}-\dfrac{3}{7}}{\dfrac{9}{11}+\dfrac{9}{3}-\dfrac{9}{7}}-\dfrac{\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{8}}{\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}+\dfrac{7}{16}}\)

\(=\dfrac{1}{3}-1:\dfrac{7}{2}=\dfrac{1}{3}-\dfrac{2}{7}=\dfrac{1}{21}\)

8 tháng 4 2023

\(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{8}+0,7}\\ =\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\\ =\dfrac{2}{7}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\\ =\dfrac{2}{7}-\dfrac{2}{7}=0\)

8 tháng 4 2023

phân số cuối là \(\dfrac{2}{7}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\) nha :vv

7 tháng 4 2022

\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\times\dfrac{2022}{2021}\)

\(M=\left(\dfrac{\dfrac{178}{495}}{\dfrac{623}{495}}-\dfrac{\dfrac{17}{60}}{\dfrac{119}{120}}\right)\times\dfrac{2022}{2021}\)

\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\times\dfrac{2022}{2021}\)

\(M=0\times\dfrac{2022}{2021}\)

M=0

27 tháng 2 2019

\(M=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right).\dfrac{2013}{2012}\)

\(M=0.\dfrac{2013}{2012}\)

\(M=0\)

\(Q=2002:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{\dfrac{-7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)

\(=2002:\left(\dfrac{1}{7}\cdot\dfrac{\dfrac{-7}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right)\)

\(=2002:\left(\dfrac{1}{7}\cdot\dfrac{-7}{2}\right)=2002:\dfrac{-1}{2}=-4004\)

21 tháng 5 2017

\(M=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right):\dfrac{2014}{2015}\)

\(=\left[\dfrac{2\left(0,2-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(0,2-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{2\left(\dfrac{1}{6}-0,125+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{7}-0,125+\dfrac{1}{10}\right)}\right]:\dfrac{2014}{2015}\)

\(=\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2014}{2015}=0:\dfrac{2014}{2015}=0\)

Vậy M = 0

21 tháng 5 2017

ko có dấu là nhân chi

12 tháng 11 2021

1.
a)10/7
b) 1
c) 3
d) 3/4
e) -1
2.
a)-3/8
b)x= 3 và x=-2
c)x=10 và x=-20

12 tháng 11 2021

Cảm ơn bạn ! ^^

20 tháng 12 2022

a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)

b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)

c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)

d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)

18 tháng 6 2018

1)

\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{6}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
\(=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)
\(=\dfrac{3}{-5}+\dfrac{3}{5}\)
\(=-\dfrac{3}{5}+\dfrac{3}{5}\)
\(=0\)

\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)

=1/64