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a) \(\dfrac{32}{12}=\dfrac{4.8}{4.3}=\dfrac{8}{3}\)
b) \(\dfrac{-26}{156}=\dfrac{-26}{26.6}=\dfrac{-1}{6}\)
c) \(\dfrac{4.7}{9.32}=\dfrac{4.7}{9.4.8}=\dfrac{7}{9.8}=\dfrac{7}{72}\)
d) \(\dfrac{2.5.13}{26.35}\dfrac{2.5.13}{2.13.5.7}=\dfrac{1}{7}\)
e)\(\dfrac{7.6-9.13}{18}=\dfrac{42-117}{18}=\dfrac{-75}{18}=\dfrac{-25.3}{6.3}=\dfrac{-25}{6}\)
f) \(\dfrac{17.5-17}{3-20}=\dfrac{17.\left(5-1\right)}{-17}=-4\)
g) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(1+7\right)}{49}=8\)
a, \(\dfrac{32}{12}=\dfrac{8}{3}\)
b, \(\dfrac{-26}{156}=\dfrac{-1}{6}\)
c, \(\dfrac{4.7}{9.32}=\dfrac{4.7}{9.4.8}=\dfrac{7}{72}\)
d,\(\dfrac{2.5.13}{26.35}=\dfrac{2.5.13}{2.13.7.5}=\dfrac{1}{7}\)
e,\(\dfrac{7.6-9.13}{18}=\dfrac{3\left(7.2-3.13\right)}{18}=\dfrac{-25}{6}\)
f ,\(\dfrac{17.5-17}{3-20}=\dfrac{17\left(5-1\right)}{-17}=-4\)
g,\(\dfrac{49+7.49}{49}=\dfrac{49\left(1+7\right)}{49}=8\)
Giải:
Ta có:
\(\dfrac{a}{b}=\dfrac{-8}{11}\left(1\right)\Leftrightarrow1-\dfrac{a}{b}=1-\dfrac{-8}{11}\)
Hay \(\dfrac{b-a}{b}=\dfrac{11+8}{11}=\dfrac{19}{11}\left(2\right)\)
Thay \(b-a=190\) vào \(\left(2\right)\) ta được:
\(\dfrac{190}{b}=\dfrac{19}{11}\Leftrightarrow190.11=19b\Leftrightarrow b=110\)
Thay \(b=110\) vào \(\left(1\right)\) ta được:
\(\dfrac{a}{110}=\dfrac{-8}{11}\Leftrightarrow11a=-8.110\Leftrightarrow a=-80\)
Vậy phân số \(\dfrac{a}{b}\) cần tìm là \(\dfrac{-80}{110}\)
Thay b - a = 190 vào (1) ta được: 
Phân số a/b phải tìm là -80/110
a) \(\dfrac{11}{21}+\dfrac{-4}{7}=\dfrac{11}{21}+\dfrac{-12}{21}=\dfrac{-1}{21}\)
b) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}=\dfrac{1}{3}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}=-1+1+\dfrac{2}{7}=\dfrac{2}{7}\)
c) \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{14}=\dfrac{28}{42}+\dfrac{30}{42}-\dfrac{9}{42}=\dfrac{49}{42}=\dfrac{7}{6}\)
d) \(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}=\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}=\dfrac{14}{35}-\dfrac{15}{35}+\dfrac{7}{35}=\dfrac{6}{35}\)
e) \(\dfrac{21}{47}+\dfrac{9}{45}+\dfrac{26}{47}+\dfrac{45}{5}=\dfrac{21}{47}+\dfrac{1}{5}+\dfrac{26}{47}+\dfrac{45}{5}=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{1}{5}+\dfrac{45}{5}\right)\)
\(=1+\dfrac{46}{5}=\dfrac{51}{5}\)
f) \(\dfrac{15}{12}-\dfrac{18}{13}+\dfrac{5}{13}-\dfrac{3}{12}=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(-\dfrac{18}{13}+\dfrac{5}{13}\right)=1+\left(-1\right)=0\)
g) \(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-4}{9}-\dfrac{5}{9}=\dfrac{-9}{9}=-1\)
h)\(\dfrac{3}{7}+\dfrac{-5}{2}-\dfrac{3}{5}=\dfrac{30}{70}+\dfrac{-175}{70}-\dfrac{42}{70}=\dfrac{-187}{70}\)
i) \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{11.16.3}{12.33.5}=\dfrac{4}{15}\)
\(a,\dfrac{7}{35},\dfrac{18}{54},\dfrac{-15}{125},\dfrac{-4}{25}\)
Các thừa số đã tối giản : \(\dfrac{-4}{25}\)
\(\dfrac{7}{35}=\dfrac{7:7}{35:7}=\dfrac{1}{5}\) , \(\dfrac{18}{54}=\dfrac{18:18}{54:18}=\dfrac{1}{3}\)
\(\dfrac{-15}{125}=\dfrac{-15:5}{125:5}=\dfrac{-3}{25}\)
\(b,\dfrac{27}{45},\dfrac{21}{28},\dfrac{8}{14},\dfrac{18}{-60},\dfrac{-270}{360}\)
Các thừa số đã tối giản là : ko có
\(\dfrac{27}{45}=\dfrac{27:9}{45:9}=\dfrac{3}{5}\) , \(\dfrac{21}{28}=\dfrac{21:7}{28:7}=\dfrac{3}{4}\)
\(\dfrac{8}{14}\)\(=\dfrac{8:2}{14:2}=\dfrac{4}{7}\) , \(\dfrac{18}{-60}=\dfrac{18:6}{-60:6}=\dfrac{3}{-10}=\dfrac{-3}{10}\)
\(\dfrac{-270}{360}=\dfrac{-270:90}{360:90}=\dfrac{-3}{4}\)
\(c,\dfrac{3.4+3.7}{6.5+9}\) = \(\dfrac{3.\left(4+7\right)}{30+9}\) = \(\dfrac{3.11}{39}\) = \(\dfrac{3.11}{3.13}=\dfrac{11}{13}\)
\(\dfrac{-63}{81},\dfrac{9.6}{9.35},\dfrac{7.2+8}{2.14.5}\)
Các p/s đã tối giản : ko có
\(\dfrac{-63}{81}=\dfrac{-63:9}{81:9}=\dfrac{-7}{9}\) , \(\dfrac{9.6}{9.35}=\dfrac{6}{35}\)
\(\dfrac{7.2+8}{2.14.5}=\dfrac{14+8}{28.5}=\dfrac{22}{140}=\dfrac{11}{70}\)
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(-\dfrac{1}{11}+\dfrac{1}{11}\right)+\left(-\dfrac{1}{12}+\dfrac{1}{12}\right)+\left(-\dfrac{1}{13}+\dfrac{1}{13}\right)+\left(-\dfrac{1}{14}+\dfrac{1}{14}\right)+\left(-\dfrac{1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\\ =\dfrac{1}{16}\)
Tính nhanh :
\(\dfrac{1}{10}+\dfrac{-1}{11}+\dfrac{1}{12}+\dfrac{-1}{13}+\dfrac{1}{14}+\dfrac{-1}{15}+\dfrac{1}{16}+\dfrac{-1}{10}+\dfrac{1}{11}+\dfrac{-1}{12}+\dfrac{1}{13}+\dfrac{-1}{14}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(\dfrac{-1}{11}+\dfrac{1}{11}\right)+\left(\dfrac{1}{12}+\dfrac{-1}{12}\right)+\left(\dfrac{-1}{13}+\dfrac{1}{13}\right)+\left(\dfrac{1}{14}+\dfrac{-1}{14}\right)\)
\(+\left(\dfrac{-1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\)
\(=0+0+...+0+\dfrac{1}{16}\)
\(=\dfrac{1}{16}\)
a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)
b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)
c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)
d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)
lưu ý mk ko chép đầu bài
mình cần gấp lắm đến chiều mai là phải nộp rùi
giúp mình nha thanks cá bạn trước 
ko có tâm trạng mà cười nữa
\(\dfrac{-270}{450}=\dfrac{-9}{15}\)
\(\dfrac{11}{-143}=\dfrac{1}{-13}\)
\(\dfrac{32}{12}=\dfrac{8}{3}\)
\(\dfrac{-26}{-156}=\dfrac{1}{6}\)