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a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)
\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)
\(=4-\frac{4}{7}=\frac{24}{7}\)
b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)
\(=8+\frac{7}{11}=\frac{95}{11}\)
c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)
\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)
\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)
\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)
d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)
\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)
\(=5\cdot\frac{7}{35}=1\)
e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)
\(=\frac{42}{43}\)
\(9\frac{2}{9}+\frac{2}{3}+7\frac{7}{9}\)
\(=9+\frac{2}{9}+\frac{2}{3}+7+\frac{7}{9}\)
\(=\left(9+7\right)+\left(\frac{2}{9}+\frac{7}{9}\right)+\frac{2}{3}\)
\(=16+1+\frac{2}{3}\)
\(=17+\frac{2}{3}\)
\(=\frac{51}{3}+\frac{2}{3}\)
\(=\frac{53}{3}\)
\(\frac{5}{9}.\frac{10}{11}+\frac{5}{9}.\frac{14}{11}-\frac{5}{9}.\frac{15}{11}\)
\(=\frac{5}{9}\left(\frac{10}{11}+\frac{14}{11}-\frac{15}{11}\right)\)
\(=\frac{5}{9}.\frac{9}{11}\)
\(=\frac{5}{11}\)
\(a,\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{3\left[\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right]}{11\left[\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right]}=\frac{3}{11}\)
Câu b tương tự
#)Giải :
b)\(\frac{5}{9}:\left(\frac{1}{3}+\frac{1}{4}\right)+\frac{5}{9}:\left(\frac{1}{9}+\frac{2}{3}\right)\)
\(=\frac{5}{9}:\frac{1}{3}+\frac{5}{9}:\frac{1}{4}+\frac{5}{9}:\frac{1}{9}+\frac{5}{9}:\frac{2}{3}\)
\(=\frac{5}{9}:\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{9}+\frac{2}{3}\right)\)
\(=\frac{5}{9}:\frac{49}{36}\)
\(=\frac{20}{49}\)
\(A=21\frac{4}{11}-\left(1\frac{3}{5}+7\frac{4}{11}\right)\)
\(A=\frac{235}{11}-\left(\frac{8}{5}+\frac{81}{11}\right)\)
\(A=\left(\frac{235}{11}-\frac{81}{11}\right)+\frac{8}{5}\)
\(A=\frac{154}{11}+\frac{8}{5}\)
\(\Rightarrow A=\frac{78}{5}\)
\(B=\left(7\frac{8}{9}+2\frac{3}{13}\right)-\left(4\frac{8}{9}-7\frac{10}{13}\right)\)
\(B=\left(\frac{71}{9}+\frac{29}{13}\right)-\left(\frac{44}{9}-\frac{101}{13}\right)\)
\(B=\left(\frac{71}{9}-\frac{44}{9}\right)+\left(\frac{29}{13}-\frac{101}{13}\right)\)
\(B=\frac{27}{9}+\frac{-72}{13}\)
\(B=3+\frac{-72}{13}\)
\(\Rightarrow B=\frac{-33}{13}\)
P/s: Hoq chắc :v
\(a)15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)
\(=15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\)
\(=15\frac{3}{13}-8\frac{3}{13}-3\frac{4}{7}\)
\(=7-3\frac{4}{7}\)
\(=6\frac{7}{7}-3\frac{4}{7}=3\frac{3}{7}\)
\(b)\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=7\frac{4}{9}-4\frac{7}{11}-3\frac{4}{9}\)
\(=7\frac{4}{9}-3\frac{4}{9}-4\frac{7}{11}\)
\(=4-4\frac{7}{11}=\frac{7}{11}\)
Đặt tử = A.Ta có:
\(A=0,75+\frac{3}{7}+\frac{3}{11}=\frac{3}{4}+\frac{3}{7}+\frac{3}{11}\)
\(=3\left(\frac{1}{4}+\frac{1}{7}+\frac{1}{11}\right)\)
Đặt mẫu = B.Ta có:
\(B=2\frac{1}{4}+\frac{9}{11}+\frac{9}{7}=\frac{9}{4}+\frac{9}{11}+\frac{9}{7}\)
\(=9\left(\frac{1}{4}+\frac{1}{11}+\frac{1}{7}\right)\)
Thay A và B vào ta đc: \(\frac{A}{B}=\frac{3\left(\frac{1}{4}+\frac{1}{7}+\frac{1}{11}\right)}{9\left(\frac{1}{4}+\frac{1}{11}+\frac{1}{7}\right)}=\frac{3}{9}=\frac{1}{3}\)