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\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)
Bài 1 :
a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)
b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)
\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)
\(=-1+\left(-2\right)=-1-2=-3\)
c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
Bài 2 :
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)
=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)
b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)
=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)
c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>
\(a.\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=-\dfrac{25}{6}\)
\(b.-\dfrac{1}{12}-\dfrac{5}{4}+\dfrac{1}{3}=-\dfrac{1}{12}-\dfrac{15}{12}+\dfrac{4}{12}=-1\)
a) \(\dfrac{-7}{2}+\dfrac{3}{4}-\dfrac{17}{12}=\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=\dfrac{-42+9-17}{12}=-\dfrac{50}{12}=-\dfrac{25}{6}\)
b)\(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)=-\dfrac{2}{24}-\left(\dfrac{63}{24}-\dfrac{8}{24}\right)\)
\(=-\dfrac{2}{24}-\dfrac{63}{24}+\dfrac{8}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)
\(\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\left(\dfrac{1}{12}+\dfrac{2}{4}\right)\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\left(\dfrac{1}{12}+\dfrac{6}{12}\right)\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\dfrac{7}{12}\right]\\ =\dfrac{3}{4}-\left[\left(-\dfrac{36}{60}\right)-\dfrac{35}{60}\right]\\ =\dfrac{3}{4}-\left(-\dfrac{71}{60}\right)\\ =\dfrac{3}{4}+\dfrac{71}{60}\\ =\dfrac{35}{60}+\dfrac{71}{60}\\ =\dfrac{106}{60}\\ =\dfrac{53}{30}\)
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\(-\dfrac{6}{7}.\dfrac{21}{12}\\ =-\dfrac{126}{84}\\ =-\dfrac{63}{42}\\ =-\dfrac{31}{14}\)
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\(\left(-5\right).\left(-\dfrac{6}{20}\right)\\ =\dfrac{\left(-5\right).\left(-6\right)}{20}\\ =\dfrac{30}{20}\\ =\dfrac{3}{2}\)
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a: =8263/19980
b: =31/99
c: \(=-\dfrac{2410}{999}\)
d: \(=\dfrac{3713}{9000}\)
e: =173/55
f: =-10771/3330
\(=-\left(\dfrac{4}{13}\cdot\dfrac{5}{17}+\dfrac{12}{13}\cdot\dfrac{4}{17}+\dfrac{4}{13}\right)\)
=-4/13
\(B=\left(\frac{3}{17}-\frac{2}{345}+\frac{5}{12}\right)-\left(-\frac{2}{345}+\frac{3}{17}-\frac{1}{12}\right)\)
\(B=\frac{3}{17}-\frac{2}{345}+\frac{5}{12}+\frac{2}{345}-\frac{3}{17}+\frac{1}{12}\)
\(B=\left(\frac{3}{17}-\frac{3}{17}\right)+\left(-\frac{2}{345}+\frac{2}{345}\right)+\frac{5}{12}+\frac{1}{12}=\frac{1}{2}\)