\(\left(2x-1\right)\left(x+\frac{2}{3}\right)=0\)
\(\orbr{\begin{cases}2x-1=0\\x+\frac{2}{3}=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-2}{3}\end{cases}}\)
\(\frac{3}{3}-\frac{2x}{-1}=\frac{2}{5}\)
\(1+2x=\frac{2}{5}\)
\(2x=\frac{-3}{5}\)
\(x=\frac{-3}{10}\)
Bài làm :
a) Ta có :
\(\left(2x-1\right)\left(x+\frac{2}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+\frac{2}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{3}\end{cases}}}\)
b) Ta có :
\(3.\left|3-2x\right|-1=\frac{2}{5}\)
\(\Leftrightarrow3.\left|3-2x\right|=\frac{2}{5}+1\)
\(\Leftrightarrow3.\left|3-2x\right|=\frac{7}{5}\)
\(\Leftrightarrow\left|3-2x\right|=\frac{7}{5}\div3\)
\(\Leftrightarrow\left|3-2x\right|=\frac{7}{15}\)
\(\Leftrightarrow\orbr{\begin{cases}3-2x=\frac{7}{15}\\3-2x=-\frac{7}{15}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{19}{15}\\x=\frac{26}{15}\end{cases}}}\)
a) \(\left(2x-1\right)\left(x+\frac{2}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+\frac{2}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-2}{3}\end{cases}}\)
b) \(3\left|3-2x\right|-1=\frac{2}{5}\)
\(\Leftrightarrow3\left|3-2x\right|=\frac{7}{5}\)
\(\Leftrightarrow\left|3-2x\right|=\frac{7}{15}\)
Còn lại tự làm nốt nha
Bài giải
a, \(\left(2x-1\right)\left(x+\frac{2}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x+\frac{2}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=1\\x=-\frac{2}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{3}\end{cases}}\)
b, \(3\left|3-2x\right|-1=\frac{2}{5}\)
\(3\left|3-2x\right|=\frac{7}{5}\)
\(\left|3-2x\right|=\frac{7}{15}\)
\(\Rightarrow\orbr{\begin{cases}3-2x=-\frac{7}{15}\\3-2x=\frac{7}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{52}{15}\\2x=\frac{38}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{26}{15}\\x=\frac{19}{15}\end{cases}}\)
