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Câu 1 :
\(a,2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\Rightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)
\(\Rightarrow7x=\frac{3}{2}-\frac{4}{5}\)
\(\Rightarrow7x=\frac{7}{10}\)\(\Leftrightarrow x=0,1\)
\(b,\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
\(\Rightarrow11x=\frac{2}{3}+1-\frac{3}{2}\)
\(\Rightarrow11x=\frac{4+6-9}{6}-\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{66}\)
Câu 2 :
\(a,\frac{2}{x-1}< 0\)
Vì \(2>0\Rightarrow\)để \(\frac{2}{x-1}< 0\)thì \(x-1< 0\Leftrightarrow x< 1\)
\(b,\frac{-5}{x-1}< 0\)
Vì \(-5< 0\)\(\Rightarrow\)để \(\frac{-5}{x-1}< 0\)thì \(x-1>0\Rightarrow x>1\)
\(c,\frac{7}{x-6}>0\)
Vì \(7>0\Rightarrow\)để \(\frac{7}{x-6}>0\)thì \(x-6>0\Rightarrow x>6\)
( x - 3/2 ) ( 2x + 1 ) > 0
TH1 : cả 2 thừa số đều lớn hơn 0
\(\Rightarrow\hept{\begin{cases}x-\frac{3}{2}>0\\2x+1>0\end{cases}\Rightarrow\hept{\begin{cases}x>\frac{3}{2}\\x>-\frac{1}{2}\end{cases}\Rightarrow}x>\frac{3}{2}}\)
TH2 : cả 2 thừa số đều bé hơn 0
\(\Rightarrow\hept{\begin{cases}x-\frac{3}{2}< 0\\2x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x< \frac{3}{2}\\x< -\frac{1}{2}\end{cases}\Rightarrow}x< -\frac{1}{2}}\)
Vậy,..........
a) \(-5x+\frac{1}{2}=\frac{2}{3}\\ -5x=\frac{2}{3}-\frac{1}{2}\\ -5x=\frac{1}{6}\\ x=\frac{1}{6}:\left(-5\right)\\ x=\frac{-1}{30}\)Vậy \(x=\frac{-1}{30}\)
b) \(\frac{1}{-5}-\frac{2}{3}+1\frac{1}{2}x=\frac{1}{2}\\ \frac{-13}{15}+\frac{3}{2}x=\frac{1}{2}\\ \frac{3}{2}x=\frac{1}{2}-\frac{-13}{15}\\ \frac{3}{2}x=\frac{41}{30}\\ x=\frac{41}{30}:\frac{3}{2}\\ x=\frac{41}{45}\)Vậy \(x=\frac{41}{45}\)
c) \(2\left(\frac{1}{4}-3x\right)=\frac{1}{5}-4x\\ \frac{1}{2}-6x=\frac{1}{5}-4x\\ \frac{1}{2}-\frac{1}{5}=6x-4x\\ \frac{3}{10}=2x\\ x=\frac{3}{10}:2\\ x=\frac{3}{20}\)Vậy \(x=\frac{3}{20}\)
d) \(\frac{-5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\\ \frac{-5}{2}-1+3x=\frac{1}{4}-7x\\ \frac{-7}{2}+3x=\frac{1}{4}-7x\\ 7x+3x=\frac{1}{4}+\frac{7}{2}\\ 10x=\frac{15}{4}\\ x=\frac{15}{4}:10\\ x=\frac{3}{8}\)Vậy \(x=\frac{3}{8}\)
a) \(\left|x+2\right|>7\)
\(\Leftrightarrow\orbr{\begin{cases}x+2>7\\x+2< -7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>5\\x< -9\end{cases}}\Leftrightarrow5< x< -9\left(ktm\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x+2< 7\\x+2>-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 5\\x>-9\end{cases}}\Leftrightarrow-9< x< 5\left(tm\right)\)
vậy....
v) \(\left|x-1\right|< 3\)
\(\Leftrightarrow\orbr{\begin{cases}x-1< 3\\x-1>-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow-2< x< 4\)
vậy...