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dễ mk bn cho mình hỏi nhé câu 4 là \(\frac{1}{2\cdot3}\)hay là\(\frac{1}{2}\cdot3\)
\(\dfrac{7}{-25}+\dfrac{18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
=\(\left(\dfrac{-7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
= \(\dfrac{11}{25} +1+\dfrac{5}{7}\)
= \(1\dfrac{11}{25}+\dfrac{5}{7}\)
= \(2\dfrac{27}{175}\)
b) \(-2+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{15}{23}+\dfrac{4}{19}\)
=\(-2+\left(\dfrac{ }{ }\right)\)
Câu 1:
A = \(\dfrac{-7}{12}+\dfrac{11}{8}-\dfrac{5}{9}=\dfrac{-42}{72}+\dfrac{99}{72}-\dfrac{40}{72}=\dfrac{-42+99-40}{72}=\dfrac{17}{72}\)
\(B=\dfrac{1}{7}-\dfrac{8}{7}:8-3:\dfrac{3}{4}.2^2=\dfrac{1}{7}-\dfrac{8}{7}.\dfrac{1}{8}-3.\dfrac{4}{3}.4=\dfrac{1}{7}-\dfrac{1}{7}-16=0-16=-16\)
\(C=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}=\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}=\dfrac{3}{7}-\dfrac{2}{3}=\dfrac{9-14}{21}=-\dfrac{5}{21}\)
Vậy A=\(\dfrac{17}{72};B=-16;C=\dfrac{-5}{21}\)
Câu 2:
a. \(\dfrac{-11x}{12}+\dfrac{3}{4}=-\dfrac{1}{6}\)
\(\Rightarrow\dfrac{-11x}{12}=-\dfrac{1}{6}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{-11x}{12}=\dfrac{-2}{12}-\dfrac{9}{12}\)
\(\Rightarrow\dfrac{-11x}{12}=\dfrac{-11}{12}\)
\(\Rightarrow-11x=\dfrac{-11.12}{12}\)
\(\Rightarrow-11x=-11\Rightarrow x=1\)
Vậy x=1
b. \(3-(\dfrac{1}{6}-x).\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow3-\left(\dfrac{1}{6}-x\right)=1\)
\(\Rightarrow-(\dfrac{1}{6}-x)=1-3\Rightarrow\dfrac{1}{6}+x=-2\)
\(\Rightarrow x=2-\dfrac{1}{6}\Rightarrow x=\dfrac{11}{6}\)
Vậy x = \(\dfrac{11}{6}\)
Câu 4:
Ta có: \(\dfrac{1}{2.3}=\dfrac{1}{6}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
A = ( -4/5 + 4/3 ) + (-5/4 + 14/5) - 7/3
= 8/15 + 31/20 - 7/3
= 25/12 - 7/3
= -1/4
B = 8/3 x 2/5 x 3/8 x 10x 19/92
= 16/15 x 15/4 x 19/92
= 4x19/92
= 19/23
C = - \(\dfrac{5}{7}\) x \(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\) x \(\dfrac{9}{14}\) + \(\dfrac{1}{57}\)
= - \(\dfrac{10}{77}\) - \(\dfrac{45}{98}\) + \(\dfrac{1}{57}\)
= - \(\dfrac{635}{1078}\) + \(\dfrac{1}{57}\)
= - \(\dfrac{36195}{61446}\) + \(\dfrac{1078}{61446}\)
= - \(\dfrac{35117}{61446}\)
\(\frac{4}{5}\)+ \(\frac{1}{5}\): \(\frac{-2}{15}\)= \(\frac{4}{5}\)+ \(\frac{1}{5}\)* \(\frac{-15}{2}\)=\(\frac{4}{5}\) + \(\frac{-3}{2}\)= \(\frac{-7}{10}\)
b = \(\frac{-7}{12}\). (\(\frac{3}{11}\)+ \(\frac{-14}{11}\)) + \(\frac{5}{6}\)= \(\frac{-7}{12}\). (-1) + \(\frac{5}{6}\)= \(\frac{7}{12}\)+ \(\frac{10}{12}\)= \(\frac{17}{12}\)
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
a)(3/4-5/6+7/12).(-12/7)=1/2.(-12/7)
=6/7
b)1,4.15/49-(4/5+2/3):11/5=1,4.15/49-22/15:11/5
=3/7-2/3=-5/21
(3/4-5/6+7/12)×(-12/7)
=(9/12-10/12+7/12)×(-12/7)
=(-1/12+7/12)×(-12/7)
=6/12×(-12/7)
=1/2×(-12/7)
=-12/14
=-6/7