Question

Tìm x

\(\dfrac{2x-3}{0.8}=\dfrac{-3.2}{3-2x}\)

Answer 1

Theo đề ta có:

\(\dfrac{2x-3}{0,8}=\dfrac{-3,2}{3-2x}\)

=> \(\left[{}\begin{matrix}2x-3.3-2x=0,8.-3,2\\3\left(2x-2x\right)=0,8.-3,2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3\left[x.\left(2-2\right)\right]=0,8.-3,2\\3\left[x.1\right]=0,8.-3,2\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}3\left[x.1\right]=-2,56\\x.1=\dfrac{-2,56}{3}\approx1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x.1=1\\\Rightarrow x=1\end{matrix}\right.\)

Vậy :...................................................