Question

[Thầy Đức Anh]

(1 điểm) Tìm $x$, biết:

a) $x-\dfrac{2}{3}=\dfrac{-5}{12}$

b) $\dfrac{8}{5}:x=\dfrac{-2}{3}$

c) $1-\dfrac{3}{7}.x=-\dfrac{2}{7}$

Answer 1

[⭐Hannie⭐]

\(x-\dfrac{2}{3}=-\dfrac{5}{12}\)

\(\Rightarrow x=-\dfrac{5}{12}+\dfrac{2}{3}\)

\(\Rightarrow x=-\dfrac{5}{12}+\dfrac{8}{12}\)

\(\Rightarrow x=\dfrac{3}{12}=\dfrac{1}{4}\)

`----`

\(\dfrac{x}{15}=\dfrac{3}{4}+\dfrac{19}{-20}\)

\(\Rightarrow\dfrac{x}{15}=-\dfrac{1}{5}\)

\(\Rightarrow-\dfrac{3}{15}=-\dfrac{1}{5}\)

\(\Rightarrow x=-3\)

`-----`

\(\dfrac{x}{-2}=-\dfrac{8}{x}\)

\(\Rightarrow x^2=-2.\left(-8\right)\)

\(\Rightarrow x^2=16\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

Answer 2

[Võ Thị Tuyết]

a) $x-\genfrac{}{}{0ex}{}{2}{3}=\genfrac{}{}{0ex}{}{-5}{12}$

$x=\genfrac{}{}{0ex}{}{-5}{12}+\genfrac{}{}{0ex}{}{2}{3}$

$x=\genfrac{}{}{0ex}{}{-5}{12}+\genfrac{}{}{0ex}{}{8}{12}$

$x=\genfrac{}{}{0ex}{}{-5+8}{12}$

$x=\genfrac{}{}{0ex}{}{3}{12}$

$x=\genfrac{}{}{0ex}{}{1}{4}$

b) $\genfrac{}{}{0ex}{}{8}{5}:x=\genfrac{}{}{0ex}{}{-2}{3}$

$x=\genfrac{}{}{0ex}{}{8}{5}:\left(\genfrac{}{}{0ex}{}{-2}{3}\right)$

$x=\genfrac{}{}{0ex}{}{8}{5}.\left(\genfrac{}{}{0ex}{}{3}{-2}\right)$

$x=\genfrac{}{}{0ex}{}{-12}{5}$

c) $1-\genfrac{}{}{0ex}{}{3}{7}.x=-\genfrac{}{}{0ex}{}{2}{7}$

$\genfrac{}{}{0ex}{}{3}{7}.x=1-\left(-\genfrac{}{}{0ex}{}{2}{7}\right)$

$\genfrac{}{}{0ex}{}{3}{7}.x=\genfrac{}{}{0ex}{}{9}{7}$

$x=\genfrac{}{}{0ex}{}{9}{7}:\genfrac{}{}{0ex}{}{3}{7}$

$x=\genfrac{}{}{0ex}{}{9}{7}.\genfrac{}{}{0ex}{}{7}{3}$

$x=3$