Question

tìm x

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

giúp mình nhé

Answer 1

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

\(\Leftrightarrow\left(4x-3\right)\cdot4=-3\cdot\left(2-3x\right)\)

\(\Leftrightarrow16x-12=-6+9x\)

\(\Leftrightarrow16x-9x=-6+12\)

\(\Leftrightarrow7x=6\Rightarrow x=\dfrac{6}{7}\)

Answer 2

\(\dfrac{4x-3}{2-3x}=\dfrac{-3}{4}\\ \Rightarrow4.\left(4x-3\right)=-3.\left(2-3x\right)\\ 16x-12=-6+9x\\ \Rightarrow16x-9x=-6+12\\ \Rightarrow7x=6\)

\(\Rightarrow x=\dfrac{6}{7}\)