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\(\Leftrightarrow\dfrac{1}{6}< \left|x-\dfrac{2}{7}\right|< \dfrac{3}{4}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-\dfrac{2}{7}\right|< \dfrac{3}{4}\\\left|x-\dfrac{2}{7}\right|>\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{13}{28}< x< \dfrac{29}{28}\\\left[{}\begin{matrix}x>\dfrac{19}{42}\\x< \dfrac{5}{42}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{13}{28}< x< \dfrac{5}{42}\\\dfrac{19}{42}< x< \dfrac{29}{28}\end{matrix}\right.\)
\(\dfrac{7}{5}+\dfrac{4}{7}-\dfrac{9}{10}=\dfrac{49-20}{35}-\dfrac{9}{10}=\dfrac{19}{35}-\dfrac{9}{10}=\dfrac{190-315}{350}=\dfrac{-125}{350}\)
\(\dfrac{2}{1}+\dfrac{3}{4}\text{×}\dfrac{8}{5}=\dfrac{8+3}{4}\text{×}\dfrac{8}{5}=\dfrac{11\text{×}8}{4\text{×}5}=\dfrac{88}{20}\)
mấy câu kia áp dụng là dc!
\(a.\frac{19}{5}\cdot\frac{4}{7}+\frac{3}{7}\cdot\frac{19}{5}-\frac{4}{5}\)
\(=\frac{19}{5}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)-\frac{4}{5}\)
\(=\frac{19}{5}\cdot1-\frac{4}{5}\)
\(=\frac{19}{5}-\frac{4}{5}=\frac{15}{5}=3\)
\(b.2\frac{2}{7}\cdot5\frac{2}{5}+\frac{16}{7}\cdot1\frac{3}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\frac{27}{5}+\frac{16}{7}\cdot\frac{8}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\left(\frac{27}{5}+\frac{8}{5}\right)+\frac{1}{2}\)
\(=\frac{16}{7}\cdot7+\frac{1}{2}\)
\(=16+\frac{1}{2}=\frac{33}{2}\)
\(c.\frac{3}{7}\cdot3\frac{3}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{15}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\left(\frac{15}{4}-\frac{5}{4}\right)-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{5}{2}-\frac{1}{4}\)
\(=\frac{15}{14}-\frac{1}{4}=\frac{23}{28}\)
Chú ý: \(\cdot:\times\)
\(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(\frac{1}{4}:x=\frac{-7}{20}\)
\(x=\frac{1}{4}:\frac{-7}{20}\)
\(x=\frac{-5}{7}\)
\(2x\left(x-\frac{1}{7}\right)=0\)
\(=>x=0\) hoặc \(x-\frac{1}{7}=0\)
\(x=0+\frac{1}{7}\)
\(x=\frac{1}{7}\)
Vậy \(x\in\left\{0;\frac{1}{7}\right\}\)
Mình làm câu b thôi nha
\(\frac{3}{4}\) + \(\frac{1}{4}\) : x = \(\frac{2}{5}\)
\(1\) : x = \(\frac{2}{5}\)
x = \(1\) * \(\frac{2}{5}\)
x = \(\frac{5}{2}\)
Answer : \(\frac{5}{2}\)
\(\dfrac{-4}{7}:x=\dfrac{-2}{5}\)
\(\Rightarrow x=\dfrac{-4}{7}:\dfrac{-2}{5}\)
\(\Rightarrow x=\dfrac{10}{7}\)
\(\dfrac{-4}{7}:x=\dfrac{-2}{5}\)
\(x=\dfrac{-4}{7}.\dfrac{-5}{2}\)
\(x=\dfrac{10}{7}\)
\(\frac{3}{2}-\left|2x-\frac{7}{4}\right|=\frac{5}{4}\)
\(\Leftrightarrow\left|2x-\frac{7}{4}\right|=\frac{3}{2}-\frac{5}{4}=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{7}{4}=\frac{1}{4}\\2x-\frac{7}{4}=-\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{4}\end{cases}}\)
Tìm \(x\)
\(\frac{3}{2}-\left|2x-\frac{7}{4}\right|=\frac{5}{4}\)
\(\left|2x-\frac{7}{4}\right|=\frac{3}{2}-\frac{5}{4}\)
\(\left|2x-\frac{7}{4}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{7}{4}=\frac{1}{4}\\2x-\frac{7}{4}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=2\\2x=1,5\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)
Vậy \(x\in\left\{0,75;1\right\}\).