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a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
e. \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}=-\frac{3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)=-\frac{3}{5}\cdot2=-\frac{6}{5}\)
f. \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{5}=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{4}{5}=\frac{1}{3}\cdot2-\frac{4}{5}=\frac{2}{3}-\frac{4}{5}=-\frac{2}{15}\)
g. \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{15}{19}+\frac{5}{7}=-\frac{3}{7}\left(\frac{4}{19}+\frac{15}{19}\right)+\frac{5}{7}=-\frac{3}{7}\cdot1+\frac{5}{7}=-\frac{3}{7}+\frac{5}{7}=\frac{2}{7}\)
h. \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
a )
\(\frac{-4}{9}.\frac{1}{3}-\frac{4}{9}.\frac{5}{6}+\frac{3}{7}.\frac{4}{9}\)
\(=\frac{4}{9}.\left(-\frac{1}{3}-\frac{5}{6}+\frac{3}{7}\right)\)
\(=\frac{4}{9}.\left(-\frac{14}{42}-\frac{35}{42}+\frac{18}{42}\right)\)
\(=\frac{4}{9}.\frac{-31}{42}\)
\(=-\frac{62}{189}\)
b )
\(\frac{2}{3}:\frac{3}{7}-\frac{2}{3}:\frac{4}{3}+\frac{2}{3}:\frac{1}{21}\)
\(=\frac{2}{3}.\frac{7}{3}-\frac{2}{3}.\frac{3}{4}+\frac{2}{3}.21\)
\(=\frac{14}{9}-\frac{1}{2}+14\)
\(=\frac{28}{18}-\frac{9}{18}+14\)
\(=\frac{19}{18}+14\)
\(=1+14+\frac{1}{18}\)
\(=15\frac{1}{18}\)
c )
\(\left(5\frac{1}{3}+3\frac{2}{3}\right)-4\frac{1}{3}\)
\(=\left(5+3-4\right)+\left(\frac{1}{3}+\frac{2}{3}-\frac{1}{3}\right)\)
\(=4\frac{2}{3}\)
\(=\frac{14}{3}\)
a) \(-\frac{4}{9}\cdot\frac{1}{3}-\frac{4}{9}\cdot\frac{5}{6}+\frac{3}{7}\cdot\frac{4}{9}\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{1}{3}+\left(-\frac{4}{9}\right)\cdot\frac{5}{6}-\left(-\frac{4}{9}\right)\cdot\frac{3}{7}\)
\(=\left(-\frac{4}{9}\right)\left(\frac{1}{3}+\frac{5}{6}-\frac{3}{7}\right)\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{31}{42}=-\frac{62}{189}\)
Bài 1:
a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)
\(=\frac{9}{15}+\frac{4}{15}\)
\(=\frac{13}{15}\)
b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)
\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)
c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)
\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)
\(=\frac{-7}{8}:\frac{-7}{4}\)
\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)
e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)
f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)
\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)
\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)
\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)
\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)
h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)
\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)
\(=\frac{5}{9}\)
dễ mà bạn đây là bài cơ bản lớp 6 dấy
câu a nhé bạn bạn nếu ko làm kiểu khó thì đổi về phaan số bình thường nà sau đó tính trong ngoặc trước rồi tính xoong bỏ dấu ngoặc nhưng ko đổi dấu né thế lad đc tương tự như các câu dưới
a)\(8\frac{2}{3}:2\frac{1}{6}-2\frac{27}{51}=\frac{26}{3}.\frac{6}{13}-\frac{43}{17}=4-\frac{43}{17}=\frac{25}{17}\)
b)\(\frac{27}{20}.\frac{15}{4}+\frac{19}{8}=\frac{119}{16}\)
c)\(\left(\frac{1}{12}+\frac{5}{6}\right)+\left(\frac{13}{35}+\frac{23}{35}\right)=\frac{11}{12}+\frac{36}{35}=\frac{817}{420}\)
d)\(\frac{24}{37}.\left(\frac{13}{18}+\frac{2}{9}+\frac{1}{18}\right)=\frac{24}{37}.1=\frac{24}{37}\)
\(4\frac{2}{5}+5\frac{6}{9}+2\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=4+\frac{2}{5}+5+\frac{6}{9}+2+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\left(4+5+2\right)+\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{6}{9}+\frac{1}{3}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)\)
\(=11+1+1+1\)
\(=14\)
\(=\frac{22}{5}+\frac{17}{3}+\frac{11}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\left(\frac{22}{5}+\frac{3}{5}\right)+\left(\frac{11}{4}+\frac{1}{4}\right)+\left(\frac{17}{3}+\frac{1}{3}\right)\)
\(=5+3+6=14\)