Question

[Thầy Đức Anh]

Thực hiện các phép tính:

a) $\dfrac{25}{6}: \dfrac{5}{3}-\left(\dfrac{-1}{4}\right);$

b) $\dfrac{1}{8} .\dfrac{3}{2}+\dfrac{1}{8} . \left(-\dfrac{1}{2}\right).$

Answer 1

[HT.Phong (9A5)]

a) \(\dfrac{25}{6}:\dfrac{5}{3}-\left(\dfrac{-1}{4}\right)\)

\(=\dfrac{25}{6}\cdot\dfrac{3}{5}+\dfrac{1}{4}\)

\(=\dfrac{5}{2}+\dfrac{1}{4}\)

\(=\dfrac{10}{4}+\dfrac{1}{4}\)

\(=\dfrac{11}{4}\)

b) \(\dfrac{1}{8}\cdot\dfrac{3}{2}+\dfrac{1}{8}\cdot\dfrac{-1}{2}\)

\(=\dfrac{1}{8}\cdot\left(\dfrac{3}{2}-\dfrac{1}{2}\right)\)

\(=\dfrac{1}{8}\cdot\dfrac{2}{2}\)

\(=\dfrac{1}{8}\)

Answer 2

[Lê Minh Long]

a) $\genfrac{}{}{0ex}{}{25}{6}:\genfrac{}{}{0ex}{}{5}{3}-\left(\genfrac{}{}{0ex}{}{-1}{4}\right)$

$=\genfrac{}{}{0ex}{}{25}{6}\cdot \genfrac{}{}{0ex}{}{3}{5}+\genfrac{}{}{0ex}{}{1}{4}$

$=\genfrac{}{}{0ex}{}{5}{2}+\genfrac{}{}{0ex}{}{1}{4}$

$=\genfrac{}{}{0ex}{}{5.2+1}{4}$

$=\genfrac{}{}{0ex}{}{11}{4}$

b) $\genfrac{}{}{0ex}{}{1}{8}\cdot \genfrac{}{}{0ex}{}{3}{2}+\genfrac{}{}{0ex}{}{1}{8}\cdot \left(-\genfrac{}{}{0ex}{}{1}{2}\right)$

$=\genfrac{}{}{0ex}{}{1}{8}\cdot \left(\genfrac{}{}{0ex}{}{3}{2}-\genfrac{}{}{0ex}{}{1}{2}\right)$

$=\genfrac{}{}{0ex}{}{1}{8}\cdot \genfrac{}{}{0ex}{}{2}{2}$

$=\genfrac{}{}{0ex}{}{1}{8}\cdot 1=\genfrac{}{}{0ex}{}{1}{8}$