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a: Ta có:
\(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)=\dfrac{2}{5}+\dfrac{2}{5}=\dfrac{4}{5}\)
\(\dfrac{4}{5}=\dfrac{4}{5}\). Vậy \(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)\)
Ta có:
\(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{7}{9}+\dfrac{1}{9}=\dfrac{8}{9}\)
\(\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{8}{9}=\dfrac{8}{9}\). Vậy \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)\)
b: \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
\(\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{1}{3}+\dfrac{6}{3}=\dfrac{7}{3}\)
\(\dfrac{7}{3}=\dfrac{7}{3}\). Vậy \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)\)
Đề của anh bị sai mới đúng chứ ạ? Anh Đạt ghi là \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}\) chứ có phải \(\dfrac{2}{5}\) đâu ạ?
M = (345 x (6789 + 3456 - 245)/690) x 99/100 x 98/99 x...x 2/3 x 1/2
M = ((345 x 10000)/690) x 99/2 (rút gọn)
M = (10000/2) x 99/2
M = 5000 x 99/2
M = 247500
Ok nha
\(1-\left(\frac{12}{5}+y=\frac{8}{9}\right):\frac{16}{9}=0\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\times\frac{16}{9}\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\)
\(\frac{12}{5}+y-\frac{8}{9}=1-0\)
\(\frac{12}{5}-y+\frac{8}{9}=1\)
\(\frac{12}{5}-y=1-\frac{8}{9}\)
\(\frac{12}{5}-y=\frac{1}{9}\)
\(y=\frac{12}{5}-\frac{1}{9}\)
\(y=\frac{108}{45}-\frac{5}{45}\)
\(y=\frac{103}{45}\)
a) $\frac{{14}}{{18}}:\frac{8}{9} = \frac{7}{9}:\frac{8}{9} = \frac{7}{9} \times \frac{9}{8} = \frac{{63}}{{72}} = \frac{7}{8}$
b) $\frac{9}{6}:\frac{3}{{10}} = \frac{3}{2}:\frac{3}{{10}} = \frac{3}{2} \times \frac{{10}}{3} = \frac{{30}}{6} = 5$
c) $\frac{4}{5}:\frac{{10}}{{15}} = \frac{4}{5}:\frac{2}{3} = \frac{4}{5} \times \frac{3}{2} = \frac{{12}}{{10}} = \frac{6}{5}$
d) $\frac{1}{6}:\frac{{21}}{9} = \frac{1}{6}:\frac{7}{3} = \frac{1}{6} \times \frac{3}{7} = \frac{3}{{42}} = \frac{1}{{14}}$
\(\left(1-\frac{2}{5}\right)\times\left(1-\frac{2}{7}\right)\times\left(1-\frac{2}{9}\right)\times\left(1-\frac{2}{11}\right)\times\left(1-\frac{2}{13}\right)\)
\(=\frac{3}{5}\times\frac{5}{7}\times\frac{7}{9}\times\frac{9}{11}\times\frac{11}{13}=\frac{3}{13}\)
\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)
\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)
\(=\frac{1}{24}\)
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)
\(=\frac{1}{10}\)
a) \(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{2}{5}\right)\times\left(1-\dfrac{2}{7}\right)\times\left(1-\dfrac{2}{9}\right)\)
\(=\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{5}{5}-\dfrac{2}{5}\right)\times\left(\dfrac{7}{7}-\dfrac{2}{7}\right)\times\left(\dfrac{9}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{2}{3}\times\dfrac{3}{5}\times\dfrac{5}{7}\times\dfrac{7}{9}\)
\(=\dfrac{2\times3\times5\times7}{3\times5\times7\times9}\)
\(=\dfrac{2}{9}\)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(=1-\dfrac{1}{9}\)
\(=\dfrac{9}{9}-\dfrac{1}{9}\)
\(=\dfrac{8}{9}\)
Sửa câu b)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
Đặt \(A=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(2A=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}\)
\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(2A=1-\dfrac{1}{9}\)
\(2A=\dfrac{9}{9}-\dfrac{1}{9}\)
\(2A=\dfrac{8}{9}\)
\(A=\dfrac{8}{9}:2\)
\(A=\dfrac{8}{18}\)
\(A=\dfrac{4}{9}\)
Vậy : \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}=\dfrac{4}{9}\)
( 1 - 1/2 ) x ( 1 - 3/5 ) x ( 1 - 2/7 ) x ( 1 - 2/9 )
= 1/2 x 2/5 x 5/7 x 7/9
= 1 x 2 x 5 x 7/2 x 5 x 7 x 9
= 1/9.