1) Tìm x:
a) \(\frac{11}{12}-\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)
\(\Leftrightarrow\frac{2}{5}+x=\frac{1}{4}:\frac{5}{12}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)
b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
\(\Leftrightarrow x=-\frac{7}{20}:\frac{1}{4}=\frac{-7}{5}\)
a) \(\frac{11}{12}-\frac{5}{12}\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{5}{12}.\frac{2}{5}-\frac{5}{12}x=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{1}{6}-\frac{5}{12}x=\frac{2}{3}\)
\(\Leftrightarrow\frac{-5}{12}x=\frac{2}{3}-\frac{11}{12}+\frac{1}{6}\)
\(\Leftrightarrow-\frac{5}{12}x=\frac{8}{12}-\frac{11}{12}+\frac{2}{12}=-\frac{1}{12}\)
\(\Leftrightarrow x=\frac{-1}{12}:\left(-\frac{5}{12}\right)=-\frac{1}{12}.\left(-\frac{12}{5}\right)=\frac{1}{5}\)
Vậy x = 1/5
b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=\frac{8}{20}-\frac{15}{20}=-\frac{7}{20}\)
\(\Leftrightarrow x=\frac{1}{4}:\left(-\frac{7}{20}\right)=\frac{1}{4}.\left(-\frac{20}{7}\right)=-\frac{5}{7}\)
Vậy x = -5/7
c) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)
d) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\end{matrix}\right.\)
Ta thấy x <-1 và x >2 vô lí
Do đó: x >-1 và x <2
Vậy -1 < x <2
e) \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy x > 2 hoặc x < -2/3
