Bài 1:
c: Ta có: \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{2}{5}=\dfrac{3}{5}\\3x-\dfrac{2}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\3x=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{15}\end{matrix}\right.\)
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