Tính một cách hợp lí:
\(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
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\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)
\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)
\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)
\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)
\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)
\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)
3) \(11\dfrac{1}{4}-\left(2\dfrac{5}{7}+5\dfrac{1}{4}\right)\)
\(=\dfrac{45}{4}-\left(7+\dfrac{27}{28}\right)\)
\(=\dfrac{45}{4}-7\dfrac{27}{28}\)
\(=\dfrac{45}{4}-\dfrac{223}{28}\)
\(=\dfrac{23}{7}\)
4) \(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{1386-14875-11250+7000-19125}{15750}\)
\(=-\dfrac{2048}{875}\)
d: =-2/7-3/11+2/7=-3/11
e: =2+3/7+1+4/7-17/7
=4-17/7=11/7
f: =-2/3*4/5+1/5*11/9
=-8/15+11/45
=-24/45+11/45=-13/45
h: =(-3,1+3,7):0,2=0,6:0,2=3
a: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}\cdot\dfrac{-17}{18}\cdot\dfrac{14}{17}=\dfrac{-14}{18\cdot9}=\dfrac{-14}{162}=\dfrac{-7}{81}\)
b: \(=\dfrac{12}{4}+\dfrac{35}{11}\cdot\dfrac{121}{245}=3+\dfrac{11}{7}=\dfrac{32}{7}\)
a: \(=\dfrac{15}{135}\cdot\dfrac{-17}{18}\cdot\dfrac{14}{17}=\dfrac{1}{9}\cdot\dfrac{-7}{9}=\dfrac{-7}{81}\)
b: \(=3+\dfrac{35}{11}\cdot\dfrac{121}{245}=3+\dfrac{11}{7}=\dfrac{32}{7}\)
\(a)\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{4}{9}\right)-\left(\dfrac{5}{7}-\dfrac{17}{14}\right)\)
\(=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{8}{18}\right)-\left(\dfrac{10}{14}-\dfrac{17}{14}\right)\)
\(=\dfrac{11}{125}-\dfrac{9}{18}-\left(-\dfrac{7}{14}\right)\)
\(=\dfrac{11}{125}-\dfrac{1}{2}+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b)1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)
\(=0+0+0-\dfrac{2}{2}-\dfrac{3}{3}-\dfrac{4}{4}\)
\(=0-1-1-1\)
\(=-3\)
a)= \(\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)+\dfrac{11}{125}\)
= \(\dfrac{-1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{11}{125}\)
= 0 + \(\dfrac{11}{125}\)
= \(\dfrac{11}{125}\)
b) \(=\left(1-1\right)+\left(\dfrac{-1}{2}-\dfrac{1}{2}\right)+\left(2-2\right)\) +
\(\left(\dfrac{-2}{3}-\dfrac{1}{3}\right)+\left(3-3\right)+\left(\dfrac{-3}{4}-\dfrac{1}{4}\right)\) + 4
= 0 + (-1) + 0 + (-1) + 0 + (-1) + 4
= -1
c) = \(\dfrac{1}{3}.\dfrac{14}{25}-\dfrac{1}{2}.\dfrac{14}{25}\)
= \(\dfrac{14}{25}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)
= \(\dfrac{14}{25}.\left(\dfrac{-1}{6}\right)\)
= \(\dfrac{-7}{75}\)
d) = \(\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\)
= 1 + (-1)
= 0
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
= 11/125 - [(17/18 - 4/9) + (5/7 - 17/14)]
= 11/125 - (1/2 - 1/2)
= 11/125
\(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{4}{9}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)=\dfrac{11}{125}-\dfrac{17-4.2}{18}+\dfrac{17-5.2}{14}=\dfrac{11}{125}-\dfrac{9}{18}+\dfrac{7}{14}=\dfrac{11}{125}-\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{11}{125}\)