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b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
\(\left(3x-\frac{2}{4}\right)\cdot\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{1}{2}\end{cases}}\)
còn lại tương tự bài trên
\(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\left[\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{5};\frac{1}{6}\right\}\)
a) \(\left(-\frac{2}{3}\right)^2:\frac{1}{3}-\left|-1\frac{1}{2}\right|=\frac{4}{9}:\frac{1}{3}-\frac{3}{2}=\frac{4}{3}-\frac{3}{2}=-\frac{1}{6}\)
b) \(\left(\frac{1}{2}-\frac{3}{5}\right)^2+\frac{2}{3}\left|\frac{3}{4}-\frac{1}{2}\right|+2012^0=\left(-\frac{1}{10}\right)^2+\frac{2}{3},\frac{1}{4}+2012^0\)
\(=\frac{1}{100}+\frac{1}{6}+1=\frac{353}{300}\)
c) \(\left(3^2:\frac{1}{3}\right)+2^3+\frac{1}{2}+\frac{1}{4}-6=3^3+2^3+\frac{3}{4}-6=29\frac{3}{4}\)
<==> ( 5x - 1)=0 < ===> 5x=1 <===>x=\(\frac{1}{5}\)
hoặc <===> (2x+\(\frac{1}{3}\)) =0 <===> 2x = \(\frac{-1}{3}\) <===> x =\(\frac{-1}{6}\)
vậy x = \(\frac{1}{5}\) hoặc x=\(\frac{-1}{6}\) nha bn !!!
bài này dễ ợt ah !!! cho mik đi :)))
\(\left(5x-1\right)\left(2x+\frac{1}{3}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}5x-1=0\\2x+\frac{1}{3}=0\end{cases}\Leftrightarrow\hept{\begin{cases}5x=1\\2x=\frac{-1}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{-1}{6}\end{cases}}}\)
\(\text{Vậy x }\varepsilon\left(\frac{1}{5};\frac{-1}{6}\right)\)
a) \(\left(2x-3\right)\left(\frac{3}{4}x+1\right)=0\)
<=>\(\hept{\begin{cases}2x-3=0\\\frac{3}{4}x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=3\\\frac{3}{4}x=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{3}{4}\end{cases}}}\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Leftrightarrow\hept{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}}\)
\(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Th1 : 5x - 1 = 0
5x = 0 + 1
5x = 1
x = 1 : 5
x = \(\frac{1}{5}\)
Th2 : 2x - \(\frac{1}{3}=0\)
2x = 0 + \(\frac{1}{3}\)
\(2x=\frac{1}{3}\)
\(x=\frac{1}{3}:2\)
\(x=\frac{1}{6}\)