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25 tháng 10 2018
Thực hiện phép tính bằng cách tính hợp lí
a)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}=\)
=\(\dfrac{15}{34}+\dfrac{1}{3}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}\)
=\(\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)-1\dfrac{15}{17}\)
=\(1+1-\dfrac{32}{17}\)
=\(\dfrac{34}{17}-\dfrac{32}{17}\)
=\(\dfrac{2}{17}\)
b)\(\left(-2\right)^3\cdot\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
=\(-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
=\(-8\cdot\dfrac{1}{2}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)\)
=\(-4:\dfrac{13}{12}\)
=\(-4\cdot\dfrac{12}{13}\)
=\(-\dfrac{48}{13}\)
8 tháng 7 2017
Thực hiện phép tính
a) \(1-\dfrac{1}{2}:\dfrac{-3}{2}+\dfrac{2}{3}=1-\dfrac{1}{2}.\dfrac{2}{-3}+\dfrac{2}{3}=\)
\(=1+\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3+1+2}{3}=\dfrac{6}{3}=2\)
b) \(\dfrac{-3}{4}-\dfrac{-7}{2}+\dfrac{2}{-3}=\dfrac{-3}{4}+\dfrac{7}{2}+\dfrac{-2}{3}=\)
\(=\dfrac{-9+42-8}{12}=\dfrac{25}{12}\)
c) \(0,25-3\dfrac{1}{2}:\dfrac{1}{2}+\dfrac{-3}{4}.\dfrac{2}{-3}=0,25-\dfrac{7}{2}.2+\dfrac{1}{2}=\)
\(=0,25-7+0,5=-\dfrac{25}{4}\)
FC
20 tháng 12 2018
\(\left(-4\right)^2.\left(\dfrac{3}{4}-0,25\right)-\left(3.\dfrac{1}{2}-1\dfrac{1}{2}\right)-\sqrt{16}\)
\(=16.\left(\dfrac{3}{4}-\dfrac{1}{4}\right)-\left(\dfrac{3}{2}-\dfrac{3}{2}\right)-4\)
\(=16.\dfrac{1}{2}-0-4\)
\(=8-0-4=8-4=4\)
14 tháng 5 2022
\(=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=36\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=\dfrac{486}{5}\)
TT
16 tháng 9 2017
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
MN
11 tháng 11 2018
e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)
=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)
=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)
=\(-\dfrac{516}{7}\)
MN
11 tháng 11 2018
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)
=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)
=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)
=\(\dfrac{119}{240}\)
19 tháng 11 2022
a: \(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{2}{17}\)
b: \(=-8\cdot\dfrac{1}{4}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-2:\dfrac{27-14}{12}=\dfrac{-2\cdot12}{13}=-\dfrac{24}{13}\)
c: \(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(44+\dfrac{4}{7}\right)\)
=-520/7
AN
7 tháng 11 2017
a)\(\dfrac{5^4.20^4}{25^4.4^5}\)
\(=\dfrac{5^4.\left(4.5\right)^4}{\left(5.5\right)^4.4^5}\)
\(=\dfrac{5^4.4^4.5^4}{5^4.5^4.4^5}\)
\(=\dfrac{4^4}{4^5}\)
\(=\dfrac{1}{4}\)
HD
7 tháng 11 2017
\(b)\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right).\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right).\dfrac{1}{2}:\dfrac{13}{12}\)
\(=\left(-4\right):\dfrac{13}{12}=\dfrac{-48}{13}\)
Chúc bạn học tốt!
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=1\\x-\dfrac{1}{3}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\left|x-\dfrac{1}{3}\right|-0,25=\dfrac{3}{4}\)
\(\left|x-\dfrac{1}{3}\right|-\dfrac{1}{4}=\dfrac{3}{4}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{3}{4}+\dfrac{1}{4}=1\)
\(\text{Vậy }x-\dfrac{1}{3}=1\)
\(x\) \(=1+\dfrac{1}{3}=\dfrac{4}{3}\)
\(\text{hoặc }x-\dfrac{1}{3}=-1\)
\(x\) \(=\left(-1\right)+\dfrac{1}{3}=\dfrac{-2}{3}\)
\(\Rightarrow x\in\left\{\dfrac{4}{3};\left(\dfrac{-2}{3}\right)\right\}\)