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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) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)
\(=\dfrac{5}{7}.-\dfrac{7}{11}\)
\(=-\dfrac{35}{77}\)
\(=-\dfrac{5}{11}\)
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
\(A=\dfrac{-3}{5}.\dfrac{3}{13}+\dfrac{-3}{5}.\dfrac{10}{13}+5\)
\(A=\dfrac{-3}{5}\left(\dfrac{3}{13}+\dfrac{10}{13}\right)+5\)
\(A=\dfrac{-3}{5}.1+5\)
\(A=\dfrac{-3}{5}+5\)
\(A=\dfrac{22}{5}\)
\(A=\left(-\dfrac{43}{51}\right)\left(-\dfrac{19}{80}\right)\)
=>A>0(1)
\(B=\left(-\dfrac{7}{13}\right)\left(-\dfrac{4}{65}\right)\left(-\dfrac{8}{21}\right)\)
=>B<0(2)
C\(=-\dfrac{5}{10}.\left(-\dfrac{4}{10}\right).....\left(\dfrac{4}{10}\right)\left(\dfrac{5}{10}\right)=0\)
=>C=0(3)
Từ 1;2;3 =>A>C>B
\(A=\dfrac{-43}{51}.\dfrac{-19}{80}\Leftrightarrow A>0\left(1\right)\)
\(B=\left(\dfrac{-7}{13}\right).\left(-\dfrac{4}{65}\right).\left(\dfrac{-8}{31}\right)\Leftrightarrow B< 0\left(2\right)\)
\(C=\dfrac{-5}{10}.\dfrac{-4}{10}...........\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\Leftrightarrow C=0\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow A>C>B\)
Ta có: \(S< \dfrac{1}{2}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{32}\) \(=\dfrac{1}{2}+\dfrac{3}{11}+\dfrac{2}{31}+\dfrac{2}{32}\)
\(=\dfrac{4909}{5456}< \dfrac{9}{10}\)
\(\Rightarrow S< \dfrac{9}{10}\)
Vậy \(S< \dfrac{9}{10}\)
\(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)
\(=\dfrac{4}{3}+\left(\dfrac{-11}{31}+\dfrac{-20}{31}\right)+\left(\dfrac{3}{10}-\dfrac{2}{5}\right)\)
\(=\dfrac{4}{3}+-1+\dfrac{-1}{10}\)
\(=\dfrac{1}{3}+\dfrac{-1}{10}\)
\(=\dfrac{7}{30}\)
Chúc bạn học tốt
= 91/93 + -107/310 - 2/5
= 19/30 - 2/5
= 7/30