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a) 7/10 - y × 3/4 = 1/5
y × 3/4 = 7/10 - 1/5
y × 3/4 = 1/2
y = 1/2 : 3/4
y = 2/3. Vậy y = 2/3
b) 5/6 : ( y + 7/9 ) = 3/4
y + 7/9 = 5/6 : 3/4
y + 7/9 = 10/9
y. = 10/9 - 7/9
y = 1/3. Vậy y = 1/3
\(\frac{7}{10}\)- y * \(\frac{3}{4}\)= \(\frac{1}{5}\)
y * \(\frac{3}{4}\)= \(\frac{7}{10}\)- \(\frac{1}{5}\)
y * \(\frac{3}{4}\)= \(\frac{1}{2}\)
y = \(\frac{1}{2}\): \(\frac{3}{4}\)
y = \(\frac{2}{3}\)
Vậy y = \(\frac{2}{3}\)
\(\frac{5}{6}\): ( y + \(\frac{7}{9}\)) = \(\frac{3}{4}\)
y + \(\frac{7}{9}\) = \(\frac{5}{6}\): \(\frac{3}{4}\)
y + \(\frac{7}{9}\) = \(\frac{10}{9}\)
y = \(\frac{10}{9}\)- \(\frac{7}{9}\)
y = \(\frac{3}{9}\)
Vậy y = \(\frac{3}{9}\)
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\(C=\dfrac{3}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{3}{7}-\dfrac{3}{7}\)
\(=\dfrac{3}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{3}{7}-\dfrac{3}{7}\times1\)
\(=\dfrac{3}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}-1\right)\)
\(=\dfrac{3}{7}\times0=0\)
\(D=8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(=4-\dfrac{31}{9}=\dfrac{36}{9}-\dfrac{31}{9}\)
\(=\dfrac{5}{9}\)
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` a) 4 8/23 + 5 7/8 - 9 8/23 `
`= 100/23 + 47/8 - 215/23 `
`= ( 100/23 - 215/23) + 47/8 `
`= -115/23 + 47/8 `
`= -5 + 47/8 `
`= -5/1 + 47/8 `
`= -40/8 + 47/8 `
`=7/8`
`b) 8 2/7 - ( 3 4/9 + 4 2/7 ) `
`= 58/7 - 31/9 - 30/7 `
`= (58/7 - 30/7 ) - 31/9 `
`= 28/7 - 31/9 `
` = 252/63 - 217/63 `
`= 35/63`
`= 5/9`
`@ animephamdanhv.`
a)\(4\dfrac{8}{23}+5\dfrac{7}{8}-9\dfrac{8}{23}=\dfrac{100}{23}+\dfrac{47}{8}-\dfrac{215}{23}=\dfrac{100}{23}-\dfrac{215}{23}+\dfrac{47}{8}=-\dfrac{115}{23}+\dfrac{47}{8}=-5+\dfrac{47}{8}=\dfrac{7}{8}\)
b) \(8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}=4-\dfrac{31}{9}=\dfrac{5}{9}\)
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3 2 phần 5 =17 phần 5 ; 2 4 phần 9=22 phần 9; 7 3 phần 8= 59 phần 8 ; 15 1 phần 10= 151 phần 10
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Bài giải
\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\)
\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{394}{90}\)
\(\frac{15}{17}-\frac{11}{13}+\frac{3}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)
\(\frac{9}{12}\text{ x }\frac{4}{3}\text{ : }\frac{8}{5}=\frac{9}{12}\text{ x }\frac{4}{3}\text{ x }\frac{5}{8}=\frac{9\text{ x }4\text{ x }5}{12\text{ x }3\text{ x }8}=\frac{5}{8}\)
\(\frac{4}{5}\text{ x }\frac{15}{8}\text{ : }\frac{5}{7}=\frac{4}{5}\text{ x }\frac{15}{8}\text{ x }\frac{7}{5}=\frac{4\text{ x }15\text{ x }7}{5\text{ x }8\text{ x }5}=\frac{21}{10}\)
\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\)
\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{197}{45}\)
\(\frac{15}{17}-\frac{11}{13}+\frac{1}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)
\(\frac{9}{12}\times\frac{4}{3}:\frac{8}{5}=1:\frac{8}{5}=\frac{5}{8}\)
\(\frac{4}{5}\times\frac{15}{8}:\frac{5}{7}=\frac{3}{2}:\frac{5}{7}=\frac{21}{10}\)
\(\frac{7}{5}+\frac{4}{7}-\frac{9}{10}=\frac{98}{70}+\frac{40}{70}-\frac{63}{70}=\frac{98+40-63}{70}=\frac{75}{70}=\frac{15}{14}\)
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