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a: \(\left(5+\frac15-\frac29\right)-\left(2-\frac{1}{23}-\frac{3}{35}+\frac56\right)-\left(8+\frac27-\frac{1}{18}\right)\)
\(=5+\frac15-\frac29-2+\frac{1}{23}+\frac{3}{35}-\frac56-8-\frac27+\frac{1}{18}\)
\(=\left(5-2-8\right)+\left(\frac15+\frac{3}{35}-\frac27\right)+\left(-\frac29-\frac56+\frac{1}{18}\right)+\frac{1}{23}\)
\(=\left(-5\right)+\left(\frac{7}{35}+\frac{3}{35}-\frac{10}{35}\right)+\left(-\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=-5+\left(-\frac{18}{18}\right)+\frac{1}{23}=-6+\frac{1}{23}=-\frac{138}{23}+\frac{1}{23}=-\frac{137}{23}\)
c: \(-\frac57-\left(-\frac{5}{67}\right)+\frac{13}{10}+\frac12+\left(-\frac16\right)+1\frac{3}{14}-\left(-\frac25\right)\)
\(=-\frac57+\frac{5}{67}+\frac{13}{10}+\frac12-\frac16+\frac{17}{14}+\frac25\)
\(=\left(-\frac57+\frac{17}{14}+\frac12\right)+\left(\frac{13}{10}+\frac25-\frac16\right)+\frac{5}{67}\)
\(=\left(-\frac{10}{14}+\frac{17}{14}+\frac12\right)+\left(\frac{13}{10}+\frac{4}{10}-\frac16\right)+\frac{5}{67}\)
\(=\left(\frac{7}{14}+\frac12\right)+\left(\frac{17}{10}-\frac16\right)+\frac{5}{67}=1+\frac{5}{67}+\frac{51}{30}-\frac{5}{30}\)
\(=\frac{72}{67}+\frac{46}{30}=\frac{72}{67}+\frac{23}{15}=\frac{2621}{1005}\)
d: \(\frac35:\left(-\frac{1}{15}-\frac16\right)+\frac35:\left(-\frac13-1\frac{1}{15}\right)\)
\(=\frac35:\left(-\frac{2}{30}-\frac{5}{30}\right)+\frac35:\left(-\frac{5}{15}-\frac{16}{15}\right)\)
\(=\frac35:\left(-\frac{7}{30}\right)+\frac35:\left(-\frac{21}{15}\right)\)
\(=\frac35\cdot\frac{-30}{7}+\frac35\cdot\frac{-5}{7}=\frac35\cdot\left(-\frac{30}{7}-\frac57\right)\)
\(=\frac35\cdot\left(-\frac{35}{7}\right)=\frac35\cdot\left(-5\right)=-3\)
a) A= 1/3 – 3/4 – (-3/5) + 1/15 – 2/9 – 1/36 b) 2/5 + (-4/3) - 1/2 c) 1/3 – [( -5/4) – (1/4 + 3/8)]
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
a)\(3-\left(\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
=\(3-\frac{1}{4}-\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
=\(\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{3}-\frac{1}{3}\right)+\left(\frac{6}{5}-\frac{3}{2}\right)\)
=\(-8+\frac{3}{2}-1-\frac{3}{10}\)
=\(\left(-8-1\right)+\left(\frac{3}{2}-\frac{3}{10}\right)\)
=-9+\(\frac{6}{5}\)
=\(\frac{-39}{5}\)
1)
a) \(-\frac{9}{34}:\frac{17}{4}\)
\(=-\frac{18}{289}.\)
b) \(1\frac{1}{2}.\frac{1}{24}\)
\(=\frac{3}{2}.\frac{1}{24}\)
\(=\frac{1}{16}.\)
c) \(-\frac{5}{2}:\frac{3}{4}\)
\(=-\frac{10}{3}.\)
d) \(4\frac{1}{5}:\left(-2\frac{4}{5}\right)\)
\(=\frac{21}{5}:\left(-\frac{14}{5}\right)\)
\(=-\frac{3}{2}.\)
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Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
a , \(\frac{7}{8}:\frac{1}{6}+\frac{7}{8}.\frac{-7}{18}\)
= \(\frac{21}{4}+\frac{-49}{144}=\frac{707}{144}\)
b, -1 : (-5) + \(\frac{1}{15}-\frac{-1}{15}\)
= \(\frac{1}{5}+0=\frac{1}{5}\)
c, \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(\frac{9}{10}-\frac{10-9}{10.9}-\frac{9-8}{9.8}-\frac{8-7}{8.7}-\frac{7-6}{7.6}-\frac{6-5}{6.5}-\frac{5-4}{5.4}-\frac{4-3}{4.3}-\frac{3-2}{3.2}.\frac{2-1}{2.1}\)
= \(\frac{9}{10}-1-\frac{1}{10}-1-\frac{1}{9}-1-\frac{1}{8}-1-\frac{1}{7}-1-\frac{1}{6}-1-\frac{1}{5}-1-\frac{1}{4}-1-\frac{1}{3}-1-\frac{1}{2}\)
= \(\frac{9}{10}-\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+...+\frac{1}{2}\right)\)
= \(\frac{9}{10}-9-1,928=\frac{9}{10}-7,071=-6.171\)
\(B=\left(\dfrac{-5}{3}+\dfrac{2}{5}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{8}{5}\right):\dfrac{4}{5}\)
\(B=\dfrac{-5}{3}+\dfrac{2}{5}:\dfrac{4}{5}+\dfrac{-1}{3}+\dfrac{8}{5}:\dfrac{4}{5}\)
\(B=\left(\dfrac{-5}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{2}{5}+\dfrac{8}{5}\right):\dfrac{4}{5}\)
\(B=\left[\left(-2\right)+2\right]:\dfrac{4}{5}\)
\(B=0:\dfrac{4}{5}=0\)
\(=>B=\dfrac{5}{4}\cdot\left(-\dfrac{5}{3}+\dfrac{2}{5}-\dfrac{1}{3}+\dfrac{8}{5}\right)=\dfrac{5}{4}\cdot\left(-2+2\right)=\dfrac{5}{4}\cdot0=0\)
\(=>C=\left(-\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{2}\right)+\left(\dfrac{1}{69}-\dfrac{1}{36}-\dfrac{2}{9}\right)\)
\(C=\left(-\dfrac{2}{6}+\dfrac{1}{6}+\dfrac{3}{6}\right)-\dfrac{65}{276}=\dfrac{1}{3}-\dfrac{65}{276}=\dfrac{157}{276}\)
=0
\(C=\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{-1}{3}+\dfrac{1}{69}-\dfrac{-2}{9}+\dfrac{1}{6}-\dfrac{1}{36}\)
\(C=\left(\dfrac{-2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{2}\right)-\left(\dfrac{1}{69}-\dfrac{1}{36}-\dfrac{2}{9}\right)\)
\(C=\dfrac{1}{3}-\dfrac{65}{276}\)
\(C=\dfrac{9}{92}\)