Ta có : \(\dfrac{2x-3}{4}=\dfrac{-36}{3-2x}\)
\(\Leftrightarrow\dfrac{2x-3}{4}=\dfrac{36}{-\left(3-3x\right)}\)
\(\Leftrightarrow\dfrac{2x-3}{4}=\dfrac{36}{2x-3}\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3\right)=4.36=144\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm12\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=12\\2x-3=-12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7,5\\x=-4,5\end{matrix}\right.\)
Vậy \(x=7,5\) hoặc \(x=-4,5\)
Tik mik nha!!!
\(\dfrac{2x-3}{4}=\dfrac{-36}{3-2x}\)
\(\Rightarrow\left(2x-3\right)\left(3-2x\right)=-36.4\)
\(\Rightarrow-\left(3-2x\right)^2=-144\)
\(\Rightarrow-\left(3-2x\right)^2=-12^2\)
\(\Rightarrow\left(3-2x\right)^2=12^2\)
\(\Rightarrow\left[{}\begin{matrix}3-2x=12\\3-2x=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-9}{2}\\x=\dfrac{15}{2}\end{matrix}\right.\).
\(\dfrac{2x-3}{4}=\dfrac{-36}{3-2x}\)
\(\Rightarrow\left(2x-3\right)\left(3-2x\right)=-36.4\)
\(\Rightarrow2x\left(3-2x\right)-3\left(3-2x\right)=-144\)
\(\Rightarrow6x-4x^2-9+4x^2=-144\)
\(\Rightarrow6x-9=-144\)
\(\Rightarrow6x=-135\)
\(\Rightarrow x=-22,5\)
mk đg cần nhưng còn nhìu bài để làm cho nên giúp mk vs nha Thanks trước