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Giải:

\(11\dfrac{3}{4}-\left(6\dfrac{5}{6}-4\dfrac{1}{2}\right)+1\dfrac{2}{3}\) 

\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{9}{2}\right)+\dfrac{5}{3}\) 

\(=\dfrac{47}{4}-\dfrac{41}{6}+\dfrac{9}{2}+\dfrac{5}{3}\) 

\(=\dfrac{141}{12}-\dfrac{82}{12}+\dfrac{54}{12}+\dfrac{20}{12}\) 

\(=\dfrac{141-82+54+20}{12}\) 

\(=\dfrac{133}{12}\) 

Chúc bạn học tốt!

26 tháng 5 2021

=133/12

d: \(\dfrac{1}{27}:\left(-\dfrac{1}{3}\right)^2+75\%\cdot\left(-\dfrac{2^2}{3}\right)\)

\(=\dfrac{1}{27}:\dfrac{1}{9}+\dfrac{3}{4}\cdot\dfrac{-4}{3}\)

\(=\dfrac{1}{3}-1\)

\(=-\dfrac{2}{3}\)

a.\(\dfrac{27}{8}\)

 

b.\(\dfrac{37}{40}\)
c.\(\dfrac{5}{2}\)

d.\(\dfrac{7}{3}\)

e.5

g.\(\dfrac{53}{16}\)

6 tháng 4 2022

Bài 1 :

a) \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{12}{8}+\dfrac{10}{8}+\dfrac{5}{8}=\dfrac{12+10+5}{8}=\dfrac{27}{8}\)

b) \(\dfrac{4}{5}-\dfrac{3}{8}+\dfrac{2}{4}=\dfrac{32}{40}-\dfrac{15}{40}+\dfrac{20}{40}=\dfrac{32-15+20}{40}=\dfrac{37}{40}\)

c) \(3+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{3}{1}+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{24}{8}+\dfrac{6}{8}-\dfrac{10}{8}=\dfrac{20}{8}=\dfrac{5}{2}\)

d) \(\dfrac{5}{6}-\dfrac{1}{2}+2=\dfrac{5}{6}-\dfrac{1}{2}+\dfrac{2}{1}=\dfrac{5}{6}-\dfrac{3}{6}+\dfrac{12}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)

e) \(\dfrac{3}{5}+\dfrac{6}{11}+\dfrac{7}{13}+\dfrac{2}{5}+\dfrac{16}{11}+\dfrac{19}{13}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)=1+2+2=5\)

g) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{29}{32}+\dfrac{13}{32}\right)=1+1+\dfrac{21}{16}=2+\dfrac{21}{16}=\dfrac{53}{16}\)

10 tháng 8 2021

Xóa nhanh đi ko CTV xóa giờ

10 tháng 8 2021

xóa j

6 tháng 7 2019

Bài làm

a)2/3-4.(1/2+3/4)=2/3-4.(2/4+3/4)

=2/3-4.5/4

=2/3-5/1

=-13/3.

b)(-1/3+5/6).11-7

=(-2/6+5/6).11-7

=3/6.11-7=1/2.11-7

=11/2-7=-3/7.

13 tháng 10 2018

\(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)

\(3^{x-1}.\left(4+2.3^3\right)=3^6.\left(4+2.3^3\right)\)

\(\Leftrightarrow3^{x-1}=3^6\)

\(\Leftrightarrow x-1=6\)

\(\Leftrightarrow x=7\)

Vậy \(x=7\)

13 tháng 10 2018

hai bài mà bn ??

18 tháng 9 2017

mk đang cần gấp người giải ai thấy bài này giải giúp mk nha 

14 tháng 6 2023

\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)

\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)

=  \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{6}{5}\)

=  \(\dfrac{1}{4}-\dfrac{6}{5}\)

=  \(-\dfrac{19}{20}\)

\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)

14 tháng 6 2023

\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)

\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)

\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)

\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)

\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)

\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)

\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)

\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)

\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)

\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)

\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)

\(=-\dfrac{5}{11}\)

\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)