14/9 x 121/7 x 3/22 =
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a: \(x-\dfrac{10}{3}=\dfrac{7}{15}\cdot\dfrac{3}{5}\)
=>\(x-\dfrac{10}{3}=\dfrac{21}{75}=\dfrac{7}{25}\)
=>\(x=\dfrac{7}{25}+\dfrac{10}{3}=\dfrac{21+250}{75}=\dfrac{271}{75}\)
b: \(x+\dfrac{3}{22}=\dfrac{27}{121}\cdot\dfrac{9}{11}\)
=>\(x+\dfrac{3}{22}=\dfrac{243}{1331}\)
=>\(x=\dfrac{243}{1331}-\dfrac{3}{22}=\dfrac{123}{2662}\)
c: \(\dfrac{8}{23}\cdot\dfrac{46}{24}-x=\dfrac{1}{3}\)
=>\(\dfrac{8}{24}\cdot\dfrac{46}{23}-x=\dfrac{1}{3}\)
=>\(\dfrac{2}{3}-x=\dfrac{1}{3}\)
=>\(x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
d: \(1-x=\dfrac{49}{65}\cdot\dfrac{5}{7}\)
=>\(1-x=\dfrac{49}{7}\cdot\dfrac{5}{65}=\dfrac{7}{13}\)
=>\(x=1-\dfrac{7}{13}=\dfrac{6}{13}\)
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Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
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\(x-\frac{10}{3}=\frac{7}{15}.\frac{3}{5}\)
\(x-\frac{10}{3}=\frac{7}{25}\)
\(x=\frac{7}{25}+\frac{10}{3}=\frac{271}{75}\)
Vậy x = 271/75
\(x-\frac{10}{3}\)\(=\)\(\frac{7}{15}\)\(.\)\(\frac{3}{5}\)
\(x-\frac{10}{3}\)\(=\) \(\frac{7}{25}\)
\(x\) \(=\)\(\frac{7}{25}\)\(+\)\(\frac{10}{3}\)
\(x\) \(=\) \(\frac{271}{75}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3}.x=\frac{1}{3}\)1) x-\(\frac{10}{3}\)=\(\frac{7}{15}.\frac{3}{5}\)
x-10/3=7/25
x=7/25+10/3
x=\(\frac{271}{75}\)
2)\(\frac{8}{23}.\frac{46}{24}.x=\frac{1}{3}\)
2/3.x=1/3
x=1/3:2/3
x=1/2
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`@` `\text {Ans}`
`\downarrow`
\(\left(\dfrac{14}{9}\div\dfrac{2}{121}\right)\times\dfrac{3}{77}\)
`=`\(\left(\dfrac{14}{9}\times\dfrac{121}{2}\right)\times\dfrac{3}{77}\)
`=`\(\left(\dfrac{7}{9}\times121\right)\times\dfrac{3}{77}\)
`=`\(\dfrac{847}{9}\times\dfrac{3}{77}=\dfrac{11}{3}\times1=\dfrac{11}{3}\)
(14/9:2/121) x 3/77
= ( 7/9 x 121/) x 3/77
= 847/9 x 3/77
=11/3 x 1=11/3
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a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
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