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![](https://rs.olm.vn/images/avt/0.png?1311)
câu đố:
có một con chim đậu trên cành 2con khác bay đến đậu.Hỏi trên cành có bao nhiêu con chim?
![](https://rs.olm.vn/images/avt/0.png?1311)
A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{1}{2}.\left(\dfrac{2}{9}+\dfrac{3}{7}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{41}{63}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\dfrac{88}{189}\)
\(=\dfrac{44}{189}\)
b) \(\left(\dfrac{-5}{28}+1,75+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\dfrac{-69}{20}\)
\(=\dfrac{-12}{23}\)
c) \(\dfrac{1}{3}.\dfrac{5}{7}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{2}{3}\)
\(=\dfrac{-3}{7}\)
d) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-\dfrac{352}{5}\)
\(=\dfrac{1}{10}\)
em không trả lời được câu hỏi của chị nhưng chị có thể giúp em đăng bài toán lên bằng cách nào không
![](https://rs.olm.vn/images/avt/0.png?1311)
c.
\(\frac{1}{3}\times\frac{5}{7}-\frac{7}{27}\times\frac{36}{14}\)
\(=\frac{1}{3}\times\frac{5}{7}-\frac{7}{3\times9}\times\frac{4\times9}{7\times2}\)
\(=\frac{1}{3}\times\frac{5}{7}-\frac{1}{3}\times\frac{4}{6}\)
\(=\frac{1}{3}\times\left(\frac{5}{7}-\frac{4}{6}\right)\)
\(=\frac{1}{3}\times\left(\frac{30-38}{42}\right)\)
\(=\frac{1}{3}\times\frac{-8}{42}\)
\(=-\frac{4}{63}\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(-\frac{3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)\)
\(=-\frac{3}{7}+\frac{15}{26}-\frac{4}{26}+\frac{3}{7}\)
\(=\left(-\frac{3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{4}{26}\right)\)
\(=0+\frac{11}{26}\)
\(=\frac{11}{26}\)
b)\(2\frac{3}{7}+\frac{2}{9}-1\frac{3}{7}\)
\(=\frac{17}{7}+\frac{2}{9}-\frac{10}{7}\)
\(=\left(\frac{17}{7}-\frac{10}{7}\right)+\frac{2}{9}\)
\(=1+\frac{2}{9}\)
\(=1\frac{2}{9}=\frac{11}{9}\)
c)\(\left(\frac{377}{-231}-\frac{124}{89}+\frac{34}{791}\right).\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{24}\right)\)
\(=\left(\frac{377}{-231}-\frac{124}{89}+\frac{34}{791}\right).\left(\frac{4}{24}-\frac{3}{24}-\frac{1}{24}\right)\)
\(=\left(\frac{377}{-231}-\frac{124}{89}+\frac{34}{791}\right).0\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B1:
a,\(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)\)
\(\Rightarrow\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(\Rightarrow\left(\frac{-3}{7}+\frac{3}{7}\right)+\frac{15}{26}-\frac{4}{26}\)
\(\Rightarrow\frac{15}{26}-\frac{4}{26}\)
=\(\frac{11}{26}\)
b,\(\frac{23}{7}+\frac{2}{9}-\frac{13}{7}\)
\(=\frac{23}{7}-\frac{13}{7}+\frac{2}{9}\)
\(=\frac{10}{7}+\frac{2}{9}\)
\(=\frac{104}{63}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}=\dfrac{15}{26}-\dfrac{4}{26}=\dfrac{11}{26}\)
b: \(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-5=\dfrac{-4}{7}-5+\dfrac{2}{9}=-\dfrac{337}{63}\)
c: \(=-\dfrac{11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =-1+17/20=-3/20
b: =(28/60-33/60)*(-25/3)
=(-1/12)*(-25/3)=1/12*25/3=25/36
c: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}=\dfrac{1}{9}\)
Giải:
a) \(\left(-\dfrac{5}{28}+1,75+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)
\(=\left(-\dfrac{5}{28}+\dfrac{7}{4}+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-\dfrac{39}{20}\right)\)
\(=\dfrac{9}{5}:\left(-\dfrac{39}{20}\right)\)
\(=\dfrac{9.\left(-20\right)}{5.39}\)
\(=\dfrac{3.\left(-4\right)}{1.13}\)
\(=\dfrac{-12}{13}\)
b) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{22}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{22}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
c) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-528.\dfrac{2}{15}\)
\(=70,5-\dfrac{1056}{15}\)
\(=70,5-70,4\)
\(=0,1\)