Question

[Hà Thảo Linh]

(Bạn nào biết thì giải giúp mình nhé, lưu ý dấu "." là dấu nhân)

Tìm x, biết:

a) -\(\dfrac{5}{6}\) - x = \(\dfrac{2}{3}\)

b) \(\dfrac{2}{3}\)x + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

c) \(\dfrac{2}{9}\) - \(\dfrac{7}{8}\) . x = \(\dfrac{1}{3}\)

d) \(\dfrac{4}{5}\) + \(\dfrac{5}{7}\) : x = \(\dfrac{1}{6}\)

e) (\(\dfrac{2}{5}\) - \(1\dfrac{2}{3}\)) : x - \(\dfrac{3}{5}\) = \(\dfrac{2}{5}\) 

f) 1 - (-\(\dfrac{3}{4}\) + x) . \(2\dfrac{2}{3}\) = 0

Answer 1

[Akai Haruma]

Lời giải:
a.

$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$

b.

$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$

$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$

c.

$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$

Answer 2

[Akai Haruma]

d.

$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$

$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$

e.

$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$

$\frac{-19}{15}: x=1$

$x=\frac{-19}{15}:1 =\frac{-19}{15}$

f.

$(-\frac{3}{4}+x).2\frac{2}{3}=1$

$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$

$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$