a) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Leftrightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=-\dfrac{1}{4}.\dfrac{20}{7}=\dfrac{-5}{7}\)
b)\(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Leftrightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{5}{-7}\)
\(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow2x=0\Rightarrow x=0\)
\(\Leftrightarrow x-\dfrac{1}{7}=0\Rightarrow x=\dfrac{1}{7}\)
b, 2x.(x-\(\dfrac{1}{7}\) )=0
x(2-\(\dfrac{1}{7}\))=0
x.\(\dfrac{13}{7}\)=0
x=0:\(\dfrac{13}{7}\)
x=0
Vậy x=0
a) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\left(-\dfrac{7}{20}\right)\)
\(\Rightarrow x=-\dfrac{5}{7}\)
b) \(2x\left(x-\dfrac{1}{7}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0:2\\x=0+\dfrac{1}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
a , \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
=> \(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
=>\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
=> x = \(\dfrac{1}{4}:\dfrac{-7}{20}\)
=> x = \(\dfrac{-5}{7}\)
b, 2x.(x- \(\dfrac{1}{7}\))=0
hoặc 2x =0 => x = 0:2 =>x =0
hoặc x-\(\dfrac{1}{7}\)=0 => x =\(\dfrac{1}{7}\)
Vậy x \(\in\)\(\left\{0;\dfrac{1}{7}\right\}\)