`@` `\text {Ans}`
`\downarrow`
\(\dfrac{2}{3}+\left[\dfrac{4}{5}x-\dfrac{11}{15}\right]=\dfrac{5}{9}\)
`=>`\(\dfrac{4}{5}x-\dfrac{11}{15}=\dfrac{5}{9}-\dfrac{2}{3}\)
`=>`\(\dfrac{4}{5}x-\dfrac{11}{15}=-\dfrac{1}{9}\)
`=>`\(\dfrac{4}{5}x=-\dfrac{1}{9}+\dfrac{11}{15}\)
`=>`\(\dfrac{4}{5}x=\dfrac{28}{45}\)
`=>`\(x=\dfrac{28}{45}\div\dfrac{ 4}{5}\)
`=>`\(x=\dfrac{7}{9}\)
Vậy, `x = 7/9.`
\(\left(\dfrac{3}{4}-3x\right)\cdot\left(1+4x\right)=0\)
`=>`\(\left[{}\begin{matrix}\dfrac{3}{4}-3x=0\\1+4x=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}3x=\dfrac{3}{4}\\4x=-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy, `x \in {1/4; -1/4}.`
