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a) Ta có : \(\frac{a}{b}+\frac{5}{6}=\frac{7}{8}\)=> \(\frac{a}{b}=\frac{7}{8}-\frac{5}{6}\)
=> \(\frac{a}{b}=\frac{21}{24}-\frac{20}{24}=\frac{1}{24}\)
Vậy \(\frac{a}{b}=\frac{1}{24}\)
b) \(\frac{a}{b}\times\frac{2}{3}=\frac{3}{4}\)
=> \(\frac{a}{b}=\frac{3}{4}:\frac{2}{3}=\frac{3}{4}\times\frac{3}{2}=\frac{9}{8}\)
Vậy \(\frac{a}{b}=\frac{9}{8}\)
c) \(\frac{5}{8}-\frac{a}{b}=\frac{5}{12}\)
=> \(\frac{a}{b}=\frac{5}{8}-\frac{5}{12}=\frac{15}{24}-\frac{10}{24}=\frac{5}{24}\)
Vậy \(\frac{a}{b}=\frac{5}{24}\)
d) \(\frac{5}{8}+\frac{a}{b}=\frac{13}{8}\)
=> \(\frac{a}{b}=\frac{13}{8}-\frac{5}{8}=\frac{8}{8}=1\)
Vậy : ....
a/b*3/5=1/5+2/3
a/b*3/5=13/15
a/b=13/15:3/5
=> a/b=13/9
Vậy phân số a/b cần tìm là 13/9
ủng hộ nha
\(\frac{a}{b}\cdot\frac{3}{5}=\frac{1}{5}+\frac{2}{3}\)
\(\Rightarrow\frac{a}{b}\cdot\frac{3}{5}=\frac{3}{15}+\frac{10}{15}\)
\(\Rightarrow\frac{a}{b}\cdot\frac{3}{5}=\frac{13}{15}\)
\(\Rightarrow\frac{a}{b}=\frac{13}{15}:\frac{3}{5}\)
\(\Rightarrow\frac{a}{b}=\frac{13}{15}\cdot\frac{5}{3}\)
\(\Rightarrow\frac{a}{b}=\frac{13}{9}\)
Vậy \(\frac{a}{b}=\frac{13}{9}\)
\(\frac{a}{b}\cdot\frac{3}{5}=\frac{1}{5}+\frac{2}{3}\)
=> \(\frac{a}{b}\cdot\frac{3}{5}=\frac{3}{15}+\frac{10}{15}=\frac{13}{15}\)
=> \(\frac{a}{b}=\frac{13}{15}:\frac{3}{5}=\frac{13}{15}\cdot\frac{5}{3}=\frac{13}{9}\)
\(\frac{A}{B}\):\(\frac{3}{5}\)=\(\frac{4}{7}\)
x =\(\frac{4}{7}\)X\(\frac{3}{5}\)
x =\(\frac{12}{35}\)
\(\frac{5}{6}\)__\(\frac{A}{B}\)=\(\frac{2}{9}\)
x =\(\frac{5}{6}\)__\(\frac{2}{9}\)
x =\(\frac{33}{54}\)
1. a
\(\dfrac{8}{5}-\dfrac{5}{6}\cdot\dfrac{3}{4}\)
\(=\dfrac{8}{5}-\dfrac{5\cdot3}{3\cdot2\cdot4}\)
\(=\dfrac{8}{5}-\dfrac{5}{8}=\dfrac{39}{40}\)
1.b
\(=\dfrac{7}{8}+\dfrac{5}{6}\cdot\dfrac{3}{2}\)
\(=\dfrac{7}{8}+\dfrac{5\cdot3}{3\cdot2\cdot2}\)
\(=\dfrac{7}{8}+\dfrac{5}{4}=\dfrac{17}{8}\)
2.a
\(\dfrac{4}{5}+x=\dfrac{11}{10}\)
\(x=\dfrac{11}{10}-\dfrac{4}{5}=\dfrac{3}{10}\)
2.b
\(x-\dfrac{3}{4}=\dfrac{5}{7}\)
\(x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{41}{28}\)
\(a+5\div b-5=\frac{3}{5}\)
\(\Rightarrow\)\(\frac{a+5}{b-5}=\frac{3}{5}\)
\(\Rightarrow\)\(\orbr{\begin{cases}a+5=3\\b-5=5\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3-5\\b=5+5\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=\left(-2\right)\\b=10\end{cases}}}\)
Vậy \(a=\left(-2\right);b=10\)