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\(\frac{\frac{4}{17}-\frac{4}{45}+\frac{4}{156}}{\frac{3}{17}-\frac{3}{45}+\frac{3}{156}}=\frac{4.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}{3.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}=\frac{4}{3}\)
Bài này về tỷ số
Khó lắm bạn ơi
Thông cảm cho mình
~~~ Chúc bạn học giỏi ~~~
B=47/41.[12.(1+1/19-1/37-1/53)/3.(1+1/19-1/37-1/53):4.(1+1/17+1/19+1/2006)/5.(1+1/17+1/19+1/2006)].123/235
=47/41.[4:4/5].123/235
=47/41.5.123/235=3
C=63.10101.37-37.10101.63/1+2+3+...+2006
=0/1+2+3+...+2006=0
CẬU XEM LẠI CHO MÌNH NHA!
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) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=5\frac{7}{7}-2\frac{5}{7}\)
\(=3\frac{2}{7}\)
Mình nghĩ đề thế này mới tính hợp lí được
2 ) B = \(1\frac{6}{41}.\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
B = 47/41 . ( 12/3 : 4/5 ) . 123/235
B = 47/41 . ( 4 : 4/5 ) . 123/235
B = 47/41 . 5 . 123/235
B = \(\frac{47.5.123}{41.235}\)
B = 3
1 ) A = \(\frac{636363.37-373737.63}{1+2+3+...+2006}\)
A = \(\frac{63.10101.37-37.10101.63}{1+2+3+...+2006}\)
A = \(\frac{0}{1+2+3+...+2006}\)
A = 0
\(A=\frac{\frac{4}{17}-\frac{4}{45}+\frac{1}{39}}{\frac{3}{17}-\frac{1}{15}+\frac{1}{52}}=\frac{\frac{4}{3}\left(\frac{4}{17}\cdot\frac{3}{4}-\frac{4}{45}\cdot\frac{3}{4}+\frac{1}{39}\cdot\frac{3}{4}\right)}{\frac{3}{17}-\frac{1}{15}+\frac{1}{52}}\)
\(A=\frac{\frac{4}{3}\left(\frac{3}{17}-\frac{1}{15}+\frac{1}{52}\right)}{\frac{3}{17}-\frac{1}{15}+\frac{1}{52}}=\frac{4}{3}\)
CHÚC BẠN HỌC TỐT!