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b: D=(-3/31-28/31)+(-6/17-11/17)+(1/25-5/25)
=-2-4/25=-54/25
\(a.\)
\(\dfrac{27}{13}-\dfrac{106}{111}+-\dfrac{5}{111}=\dfrac{27}{13}-\dfrac{106}{111}-\dfrac{5}{111}=\dfrac{27}{13}-\left(\dfrac{106+6}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)
\(b.\)
\(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
\(c.\)
\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+-\dfrac{6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)-\left(\dfrac{25}{31}+\dfrac{6}{31}\right)=1-1=0\)
a) \(\dfrac{27}{13}-\dfrac{106}{111}+\dfrac{-5}{111}=\dfrac{27}{13}+\left(\dfrac{-106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}+-1=\dfrac{14}{13}\)
b) \(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\left(\dfrac{7}{19}+\dfrac{12}{19}\right)=\dfrac{12}{11}+1=\dfrac{23}{11}\)
c)\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+\dfrac{-6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{-25}{31}+\dfrac{-6}{31}\right)=1+-1=0\)
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
1) \(\dfrac{17}{5}\cdot\dfrac{-31}{125}\cdot\dfrac{1}{2}\cdot\dfrac{10}{17}\cdot\dfrac{-1}{2^3}\)
\(=\dfrac{17\cdot\left(-31\right)\cdot1\cdot2\cdot5\cdot\left(-1\right)}{5\cdot125\cdot2\cdot17\cdot8}\)
\(=\dfrac{\left(-31\right)\left(-1\right)}{125\cdot8}\\ =\dfrac{31}{1000}\)
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1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)
\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{19}{18}+\dfrac{5}{18}\)
\(=\dfrac{24}{18}\)
\(=\dfrac{4}{3}\)
2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\dfrac{1}{15}+\dfrac{7}{15}\)
\(=\dfrac{8}{15}\)
3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)
\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
\(=\dfrac{7}{30}\)
4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)
\(=\dfrac{5}{7}.-\dfrac{7}{11}\)
\(=-\dfrac{35}{77}\)
\(=-\dfrac{5}{11}\)
Xét: \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)
\(=\dfrac{3-2-1}{6}\)
\(=0\)
\(\rightarrow C=0\)
\(a,MSC:180\\ \dfrac{17}{12}=\dfrac{17.15}{12.15}=\dfrac{255}{180};\dfrac{31}{18}=\dfrac{31.10}{18.10}=\dfrac{310}{180};\dfrac{8}{15}=\dfrac{8.12}{15.12}=\dfrac{96}{180}\\ b,MSC:75\\ \dfrac{7}{15}=\dfrac{7.5}{15.5}=\dfrac{35}{75};\dfrac{8}{25}=\dfrac{8.3}{25.3}=\dfrac{24}{75};\dfrac{11}{75}=\dfrac{11}{75}\)
=\(-\dfrac{54}{25}\)
= -24/25