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\(1-\left(\frac{12}{5}+y=\frac{8}{9}\right):\frac{16}{9}=0\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\times\frac{16}{9}\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\)
\(\frac{12}{5}+y-\frac{8}{9}=1-0\)
\(\frac{12}{5}-y+\frac{8}{9}=1\)
\(\frac{12}{5}-y=1-\frac{8}{9}\)
\(\frac{12}{5}-y=\frac{1}{9}\)
\(y=\frac{12}{5}-\frac{1}{9}\)
\(y=\frac{108}{45}-\frac{5}{45}\)
\(y=\frac{103}{45}\)
a: 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
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) $\frac{1}{6}:\frac{{21}}{9} = \frac{1}{6}:\frac{7}{3} = \frac{1}{6} \times \frac{3}{7} = \frac{3}{{42}} = \frac{1}{{14}}$
\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)
\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)
\(=\frac{1}{24}\)
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)
\(=\frac{1}{10}\)
\(x+\frac{3}{4}=\frac{5}{3}\) \(\frac{7}{5}:x=3\) \(2-x=\frac{4}{7}\)
x =\(\frac{5}{3}-\frac{3}{4}\) x=\(\frac{7}{5}:3\) x=\(2-\frac{4}{7}\)
x=\(\frac{11}{12}\) x=\(\frac{7}{15}\) x=\(\frac{10}{7}\)
3/7x2=6/7
5/12x1=5/12
3x4/7=12/7
0x5/9=0