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Bùi Hương Thảo
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Trần Tuyết Như
25 tháng 6 2015 lúc 20:21

\(\frac{\frac{2}{45}-\frac{4}{13}-\frac{1}{3}}{\frac{3}{13}-\frac{4}{15}+\frac{2}{3}}=\frac{\frac{-349}{585}}{\frac{41}{65}}=\frac{-349}{585}:\frac{41}{65}=\frac{-349}{585}\cdot\frac{65}{41}=-3\frac{52}{99}\)

Trương Thái Hậu
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Đỗ Diệu Linh
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Nguyễn Lê Phước Thịnh
12 tháng 5 2022 lúc 8:00

2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)

\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)

3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)

=2/7-2/7=0

kuyhl
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kuyhl
4 tháng 4 2016 lúc 21:32

=

\(\frac{\left(229-183\right).30}{\frac{45}{\frac{4}{3}}}-\frac{\frac{17}{4}.\frac{20}{17}+1.\frac{2}{1}}{\frac{1}{2}+\frac{13}{2}}=\frac{\frac{46}{15}.30}{\frac{4}{3}}-\frac{5+2}{7}=\frac{92}{3}.\frac{3}{4}-1=23-1=22\)

ai k mk mk k lại

thảo nguyễn thanh
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Hoàng Quân
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Hà Nguyễn Thanh Hải
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a) \(\frac{1}{12}+\frac{3}{15}+\frac{11}{12}+\frac{1}{71}-\frac{12}{10}=\left(\frac{1}{12}+\frac{11}{12}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{71}\)

\(=\frac{12}{12}+0+\frac{1}{71}=1+\frac{1}{71}=1\frac{1}{71}=\frac{72}{71}\)

b) \(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4.\frac{5}{4}=\frac{2}{3}-5=\frac{2}{3}-\frac{15}{3}=-\frac{13}{3}\)

c) \(\frac{-4}{13}.\frac{3}{17}+\frac{-12}{13}.\frac{4}{7}+\frac{4}{13}=\frac{4}{13}.\frac{-3}{17}+\frac{4}{13}.\frac{-12}{17}+\frac{4}{13}.1\)

\(=\frac{4}{13}\left(\frac{-3}{17}+\frac{-12}{17}+1\right)=\frac{4}{13}\left(\frac{-15}{17}+\frac{17}{17}\right)=\frac{4}{13}.\frac{2}{17}=\frac{8}{221}\)

d) \(\frac{10^3+2.5+5^3}{55}=\frac{1000+10+125}{55}=\frac{1135}{55}=\frac{227}{11}\)

Khách vãng lai đã xóa
Erza Scarlet
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Đặng Quỳnh Ngân
10 tháng 8 2016 lúc 15:43

a) \(\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)

\(=\left(\frac{1}{3}.\frac{-9}{10}\right)\left(\frac{-6}{13}.\frac{-13}{36}\right)\)

\(=\frac{-3}{10}.\frac{1}{6}\)

\(=\frac{-1}{20}\)

b) \(\frac{-1}{3}.\frac{-15}{17}.\frac{34}{45}\)

\(=\frac{-1}{3}.\frac{-2}{3}\)

\(=\frac{2}{9}\)

c) \(\left(1-\frac{1}{5}\right)\left(\frac{-3}{10}+\frac{1}{5}\right)\)

\(=\frac{4}{5}.\frac{-1}{10}\)

\(=\frac{-2}{25}\)

d) \(A=\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}+\frac{2}{3}\)

\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)+\frac{2}{3}\)

\(=\frac{1}{3}.2+\frac{2}{3}\)

\(=\frac{2}{3}+\frac{2}{3}\)

\(=\frac{4}{3}\)

e)  \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)

\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)

\(=6-2\frac{5}{7}\)

\(=5\frac{7}{7}-2\frac{5}{7}\)

\(=3\frac{2}{7}\)

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