Bài 2:
Câu 1:
\(PT\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{1}{6}\\x-\dfrac{1}{4}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{12};\dfrac{1}{12}\right\}\)
Câu 2:
\(PT\Leftrightarrow\left|3x-5\right|=\dfrac{10}{21}\Leftrightarrow\left[{}\begin{matrix}3x-5=\dfrac{10}{21}\\3x-5=-\dfrac{10}{21}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{115}{63}\\x=\dfrac{95}{63}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{115}{63};\dfrac{95}{63}\right\}\)
Câu 3:
\(PT\Leftrightarrow-\dfrac{16x}{15}.\dfrac{1}{7}=-\dfrac{5}{21}\Leftrightarrow-16x=-25\Leftrightarrow x=\dfrac{25}{16}\)
Vậy: \(x=\dfrac{25}{16}\)
\(1,\left(x-\dfrac{1}{4}\right)^2=\left(\dfrac{-1}{6}\right)^2=\left(\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow x-\dfrac{1}{4}=\left\{{}\begin{matrix}\dfrac{-1}{6}\\\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{6}+\dfrac{1}{4}\\x=\dfrac{1}{6}+\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{12}\\x=\dfrac{5}{12}\end{matrix}\right.\)
\(2,\left|3x-5\right|=\dfrac{1}{3}+\dfrac{1}{7}\)
\(\Leftrightarrow\left|3x-5\right|=\dfrac{10}{21}\)
\(\Leftrightarrow3x-5=\left\{{}\begin{matrix}\dfrac{10}{21}\\\dfrac{-10}{21}\end{matrix}\right.\)
\(\Leftrightarrow3x=\left\{{}\begin{matrix}\dfrac{115}{21}\\0\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{115}{63};0\)
