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\(x\in c???\)
Đề bài là : Tìm \(x\inℤ\)biết :
\(\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{2}\right]:\frac{-5}{6}< x< \frac{4}{21}:\frac{4}{-7}\)
\(\Rightarrow\left[\frac{1}{2}-\frac{1}{2}+\frac{3}{4}\right]:\frac{-5}{6}< x< \frac{4}{21}:\frac{-4}{7}\)
\(\Rightarrow\left[0+\frac{3}{4}\right]:\frac{-5}{6}< x< \frac{4}{21}\cdot\frac{7}{-4}\)
\(\Rightarrow\frac{3}{4}:\frac{-5}{6}< x< \frac{1}{3}\cdot\frac{1}{-1}\)
\(\Rightarrow\frac{3}{4}\cdot\frac{6}{-5}< x< -\frac{1}{3}\)
\(\Rightarrow\frac{-9}{10}< x< -\frac{1}{3}\)
\(\Rightarrow\frac{-27}{30}< x< -\frac{10}{30}\)
Tự tìm :v
Bài 1:
1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3
= \(\frac{11}{3}\): \(\frac{10}{3}\)- 3
= \(\frac{11}{3}\). \(\frac{3}{10}\)- 3
= \(\frac{11}{10}\)- 3
= \(\frac{-19}{10}\)
2) \(\frac{5}{6}\): \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\) . \(\frac{52}{3}\)- \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))
= \(\frac{5}{6}\).( -30)
= -25
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)< x< \frac{2}{3}.\left(\frac{-1}{6}+\frac{3}{4}\right)\)
⇒ \(\frac{4}{3}.\left(\frac{-1}{3}\right)< x< \frac{2}{3}.\left(\frac{7}{12}\right)\)
⇒ \(\frac{-4}{9}< x< \frac{7}{18}\)
⇒ \(\frac{-8}{18}< x< \frac{7}{18}\)
mà -8<x<7
⇒ x ϵ \(\left\{-7;-6;-5;-4;....;5;6\right\}\)
\(\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right):\frac{-5}{6}< x< \frac{4}{21}.\frac{4}{7}\)
\(\Rightarrow\left(\frac{6}{12}+\frac{9}{12}-\frac{4}{12}\right):\frac{-10}{12}< x< \frac{16}{147}\)
\(\Rightarrow\frac{11}{12}.\frac{-12}{10}< x< \frac{16}{147}\)
\(\Rightarrow\frac{-11}{10}< x< \frac{16}{147}\)
\(\Rightarrow\frac{-1617}{1470}< x< \frac{16}{1470}\)
\(x=\left\{-1;0\right\}\)