Question

2x.(x-\(\dfrac{1}{7}\))=0

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

Answer 1

*\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

Vậy phương trình đã cho có \(S=\left\{0;\dfrac{1}{7}\right\}\)

* \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\left(x\ne0\right)\)

\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4x}=\dfrac{2}{5}\)

\(\Leftrightarrow15x+1=8x\)

\(\Leftrightarrow7x=-1\)

\(\Leftrightarrow x=\dfrac{-1}{7}\left(tmđk\right)\)

Vậy phương trình đã cho có \(S=\left\{\dfrac{-1}{7}\right\}\)

Answer 2

a, \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

Vậy \(x\in\left\{0;\dfrac{1}{7}\right\}\)

b, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\Rightarrow x=\dfrac{-5}{7}\)

Vậy \(x=\dfrac{-5}{7}\)

Chúc bạn học tốt!!!

Answer 3

a) \(2x.\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

b) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{-5}{7}\)

Chúc bạn học tốt

Answer 4

\(2x\cdot\left(x-\dfrac{1}{7}\right)=0\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\\ \dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\\ \dfrac{1}{4}:x=\dfrac{-7}{20}\\ x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=\dfrac{-5}{7}\)

Answer 5

\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

Answer 6

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\\ \Leftrightarrow\dfrac{1}{4}.\dfrac{1}{x}=\dfrac{2}{5}-\dfrac{3}{4}\\ \Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\\ \Leftrightarrow4x=-\dfrac{20}{7}\\ \Leftrightarrow x=-\dfrac{5}{7}\)

Answer 7

Answer 8

a) \(2x.\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow2x=0\Rightarrow x=0\)

\(\Leftrightarrow x-\dfrac{1}{7}=0\Rightarrow x=\dfrac{1}{7}\)

b)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{-5}{7}\)