Question

Tìm x biết : \(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)

Answer 1

\(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\dfrac{3}{2}=\dfrac{3}{2}\\x=\dfrac{-1}{2}.\dfrac{4}{3}=\dfrac{-4}{6}=\dfrac{-2}{3}\end{matrix}\right.\)

Answer 2

\(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\dfrac{2}{3}x-1=0\Rightarrow\dfrac{2}{3}x=1\Rightarrow x=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{3}{4}x+\dfrac{1}{2}=0\Rightarrow\dfrac{3}{4}x=-\dfrac{1}{2}\Rightarrow x=-\dfrac{2}{3}\)

Vậy.....

Answer 3

Mình k chép lại đề nhé!

<=> \(\dfrac{2}{3}x-1=0\)=> \(\dfrac{2}{3}x=1=>x=\dfrac{3}{2}\)

hoặc \(\dfrac{3}{4}x+\dfrac{1}{2}=0=>\dfrac{3}{4}x=-\dfrac{1}{2}=>x=-\dfrac{2}{3}\)

Vậy x thuộc {-2/3 ; 3/2}

Answer 4

\(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}x=1\\\dfrac{3}{4}x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy...

Answer 5

\(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\\ \left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=1\\\dfrac{3}{4}x=\dfrac{-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\div\dfrac{2}{3}\\x=-\dfrac{1}{2}\div\dfrac{3}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(\text{Vậy }\) \(x=\dfrac{3}{2}\) \(\text{hoặc}\) \(x=-\dfrac{2}{3}\)