\(501\frac{1}{3}.10.\frac{7}{35}.0,75\%\)
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a, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}.\frac{12}{7}\)
\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=\frac{-5}{7}.1+\frac{12}{7}=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
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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: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)
\(=\frac{42}{43}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)
\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(A=\frac{11}{3}\)(1)
+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)
\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)
=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)
=> \(\frac{11}{3}-x=1\)
=> \(x=\frac{11}{3}-1=\frac{8}{3}\)
Vậy x = 8/3
501.1/3.10.7/35.0,75 = 501.1/3.10.1/5.75/100
= 250 / 1 /2