Question

có bạn nào giúp mk dc ko? 11h mk pk lên trường r

tìm x biết \(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\)

Answer 1

\(\Leftrightarrow\left|2x-1\right|=\dfrac{7}{2}:\dfrac{21}{22}=\dfrac{7}{2}\cdot\dfrac{22}{21}=\dfrac{11}{3}\)

=>2x-1=11/3 hoặc 2x-1=-11/3

=>2x=14/3 hoặc 2x=-8/3

=>x=7/3 hoặc x=-4/3

Answer 2

Nếu \(x>\dfrac{1}{2}\) , ta có:

\(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\Rightarrow\dfrac{7}{2}:\left(2x-1\right)=\dfrac{21}{22}\Rightarrow x=\dfrac{7}{3}\left(tm\right)\)

Nếu \(x< \dfrac{1}{2}\), ta có:

\(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\Rightarrow\dfrac{7}{2}:\left(1-2x\right)=\dfrac{21}{22}\Rightarrow-2x=\dfrac{8}{3}\Rightarrow x=-\dfrac{4}{3}\left(tm\right)\)

Vậy \(x=\dfrac{7}{3};x=\dfrac{4}{3}\)

Answer 3

\(=>\left|2x-1\right|=3\dfrac{1}{2}:\dfrac{21}{22}=\dfrac{7}{2}:\dfrac{21}{22}=\dfrac{11}{3}\)

\(=>\left[{}\begin{matrix}2x-1=\dfrac{11}{3}\\2x-1=-\dfrac{11}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{14}{3}\\2x=-\dfrac{8}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)