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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?
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) $\frac{1}{6}:\frac{{21}}{9} = \frac{1}{6}:\frac{7}{3} = \frac{1}{6} \times \frac{3}{7} = \frac{3}{{42}} = \frac{1}{{14}}$
\(a,\frac{10}{21}×\frac{14}{25}=\frac{4}{15}\)
\(b,3\frac{1}{3}×\frac{12}{35}=\frac{10}{3}×\frac{12}{35}=\frac{8}{7}\)
\(c,\frac{7}{15}×\frac{11}{13}+\frac{7}{15}×\frac{2}{13}+\frac{8}{15}=\frac{7}{15}+\frac{8}{15}=1\)
\(d,\left(\frac{41}{75}+\frac{17}{10}\right)×\frac{80}{129}=\frac{4}{9}\)
\(a.\frac{4}{3}-\frac{3}{2}:X=\frac{1}{6}\)
\(\frac{3}{2}:X=\frac{4}{3}-\frac{1}{6}\)
\(\frac{3}{2}:X=\frac{8}{6}-\frac{1}{6}\)
\(\frac{3}{2}:X=\frac{7}{6}\)
\(X=\frac{3}{2}:\frac{7}{6}\)
\(X=\frac{3}{2}\times\frac{6}{7}\)
\(X=\frac{9}{7}\)
\(b.\left(X+\frac{2}{3}\right):\frac{1}{3}=\frac{41}{3}\)
\(X-\frac{2}{3}=\frac{41}{3}.\frac{1}{3}\)
\(X-\frac{2}{3}=\frac{41}{9}\)
\(X=\frac{41}{9}+\frac{2}{3}\)
\(X=\frac{41}{9}+\frac{6}{9}\)
\(X=\frac{47}{9}\)
\(\left(2.8x-32\right):\frac{2}{3}=90\)
\(2.8\cdot x-32=90\cdot\frac{2}{3}\)
\(\frac{14}{5}x-32=60\)
\(\frac{14}{5}x=60+32\)
\(\frac{14}{5}x=92\)
\(x=\frac{230}{7}\)
B , c , d tương tự
a: =3990-(463*72-39^2):15
=3990-2121=1869
b: =1995-963-172=860
c: =33/11-7/11=26/11
d: =3/2+5=13/2
`@` `\text {Answer}`
`\downarrow`
`a,`
`3990 - (463 \times 72 - 39 \times 39) \div 15`
`= 3990 - (33336 - 1521) \div 15`
`= 3990 - 31815 \div 15`
`= 3990 - 2121`
`= 1869`
`b,`
`1995 - 321 \times 3 - 6020 \div 35`
`= 1995 - 963 - 172`
`= 1032 - 172`
`= 860`
`c,`
`3 - 7/11`
`= 33/11 - 7/11`
`= 26/11`
`d,`
`1/4 \div 1/6 + 5`
`= 1/4 \times 6 + 5`
`= 3/2 + 5`
`= 13/2`